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Exercises 1 There are many methods we can use to find the product of two binomials. Here we will review two of them — the FOIL Method and using algebra tiles.Using the FOIL Method The word FOIL is an acronym for the words First, Outer, Inner, and Last. This is a mnemonic to remind us the order to follow when multiplying binomials.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution19953_1_604487304_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b=mlg.board([-2.38,3,4.62,0],{"grid":{"visible":false},"desktopSize":"large", "style":"usa"}); var t =600;b.txt(0-2,1,"(",{fontSize:1.3}); b.txt(0.25-2,1,"a",{fontSize:1.3}); b.txt(0.5-2,1,"+",{fontSize:1.3}); b.txt(0.75-2,1,"b",{fontSize:1.3}); b.txt(1.0-2,1,")",{fontSize:1.3}); b.txt(1.25-2,1,"(",{fontSize:1.3}); b.txt(1.5-2,1,"c",{fontSize:1.3}); b.txt(1.75-2,1,"+",{fontSize:1.3}); b.txt(2.0-2,1,"d",{fontSize:1.3}); b.txt(2.25-2,1,")",{fontSize:1.3});var t1 = b.txt(1.12,2.5," \\textcolor{red}{\\text{F}}\\text{irst terms}",{"fontsize":1.3,opacity:0}); var f1 = b.bendyArrow([0.27-2,1.20],[1.48-2,1.2],40,140,{strokeColor:'red',opacity:0});var t2= b.txt(1.12,2.5," \\textcolor{#2999f0}{\\text{O}}\\text{uter terms}",{"fontsize":1.3,opacity:0}); var f2= b.bendyArrow([0.27-2,1.20],[1.98-2,1.2],50,130,{strokeColor:'#2999f0',opacity:0});var t3= b.txt(1.12,2.5," \\textcolor{#02b03e}{\\text{I}}\\text{nner terms}",{"fontsize":1.3,opacity:0}); var f3= b.bendyArrow([0.77-2,0.8],[1.48-2,0.8],-40,-140,{strokeColor:'#02b03e',opacity:0});var t4 = b.txt(1.12,2.5," \\textcolor{#a72cd4}{\\text{L}}\\text{ast terms}",{"fontsize":1.3,opacity:0}); var f4 = b.bendyArrow([0.77-2,0.8],[1.98-2,0.8],-50,-150,{strokeColor:'#a72cd4',opacity:0});var p1 = b.txt(0.85,1,"= \\ ab",{fontSize:1.3,opacity:0}); var l1 = b.txt(1.1,1.65," \\ \\textcolor{red}{\\text{F}}",{fontSize:1.3,opacity:0});var p2 =b.txt(1.75,1,"+ \\ ad",{fontSize:1.3,opacity:0}); var l2 =b.txt(1.9,1.65," \\ \\textcolor{#2999f0}{\\text{O}}",{fontSize:1.3,opacity:0});var p3 = b.txt(2.65,1,"+ \\ bc",{fontSize:1.3,opacity:0}); var l3 = b.txt(2.8,1.65," \\ \\textcolor{#02b03e}{\\text{I}}",{fontSize:1.3,opacity:0});var p4 =b.txt(3.55,1,"+ \\ bd",{fontSize:1.3,opacity:0}); var l4 =b.txt(3.7,1.65," \\ \\textcolor{#a72cd4}{\\text{L}}",{fontSize:1.3,opacity:0});mlg.af("A",function() { $('#btn1').css({'background-color':'#24c5ed'});b.hide(p1,0); b.hide(l1,0); b.hide(f1,0); b.hide(p2,0); b.hide(l2,0); b.hide(f2,0); b.hide(p3,0); b.hide(l3,0); b.hide(f3,0); b.hide(p4,0); b.hide(l4,0); b.hide(f4,0);b.show(f1,t); setTimeout( function() { b.show(t1,t); }, 2*t); setTimeout( function() { b.hide(t1,t); }, 4*t); setTimeout( function() { b.show(l1,t); }, 4*t); setTimeout( function() { b.show(p1,t); }, 4*t);setTimeout( function() { b.show(f2,t); }, 6*t); setTimeout( function() { b.show(t2,t); }, 8*t); setTimeout( function() { b.hide(t2,t); }, 10*t); setTimeout( function() { b.show(l2,t); }, 10*t); setTimeout( function() { b.show(p2,t); }, 10*t);setTimeout( function() { b.show(f3,t); }, 12*t); setTimeout( function() { b.show(t3,t); }, 14*t); setTimeout( function() { b.hide(t3,t); }, 16*t); setTimeout( function() { b.show(l3,t); }, 16*t); setTimeout( function() { b.show(p3,t); }, 16*t);setTimeout( function() { b.show(f4,t); }, 18*t); setTimeout( function() { b.show(t4,t); }, 20*t); setTimeout( function() { b.hide(t4,t); }, 22*t); setTimeout( function() { b.show(l4,t); }, 22*t); setTimeout( function() { b.show(p4,t); }, 22*t);setTimeout( function() { $('#btn1').css({'background-color':'#d8dce3'}); }, 24*t); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution19953_1_604487304_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution19953_1_604487304_l", "Solution19953_1_604487304_p", 1, code); }); } ); } window.JXQtable["Solution19953_1_604487304_l"] = true;Calculate Productwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#btn1').on('touchstart mousedown', function () { try { mlg.cf("A"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); For example, let's consider that we want to find the product of the binomials (x+2) and (x−1). We just need to multiply following the order indicated by the FOIL acronym and simplify by combining like terms.Using Algebra Tiles We can represent algebraic expressions using algebra tiles. Let's consider the same product one more time, (x+2)(x−1).The tiles in the rectangular array shown above represent each of the binomials. To find their product we need to complete the diagram first. Recall that the product of algebra tiles with the same sign gives a positive result, while those with different signs give a negative result.Now we can write the algebraic expression being represented by the algebra tiles. (x+2)(x−1)=x2−x+2x−2 Finally, we just need to simplify. (x+2)(x−1)(x+2)(x−1)=x2−x+2x−2=x2+x−2 | |

Exercises 2 Recall that the word FOIL is an acronym for the words First, Outer, Inner, and Last. This is a mnemonic to remind us of the order to follow when multiplying two binomials.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution19954_1_2063897220_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b=mlg.board([-2.38,3,4.62,0],{"grid":{"visible":false},"desktopSize":"large", "style":"usa"}); var t =600;b.txt(0-2,1,"(",{fontSize:1.3}); b.txt(0.25-2,1,"a",{fontSize:1.3}); b.txt(0.5-2,1,"+",{fontSize:1.3}); b.txt(0.75-2,1,"b",{fontSize:1.3}); b.txt(1.0-2,1,")",{fontSize:1.3}); b.txt(1.25-2,1,"(",{fontSize:1.3}); b.txt(1.5-2,1,"c",{fontSize:1.3}); b.txt(1.75-2,1,"+",{fontSize:1.3}); b.txt(2.0-2,1,"d",{fontSize:1.3}); b.txt(2.25-2,1,")",{fontSize:1.3});var t1 = b.txt(1.12,2.5," \\textcolor{red}{\\text{F}}\\text{irst terms}",{"fontsize":1.3,opacity:0}); var f1 = b.bendyArrow([0.27-2,1.20],[1.48-2,1.2],40,140,{strokeColor:'red',opacity:0});var t2= b.txt(1.12,2.5," \\textcolor{#2999f0}{\\text{O}}\\text{uter terms}",{"fontsize":1.3,opacity:0}); var f2= b.bendyArrow([0.27-2,1.20],[1.98-2,1.2],50,130,{strokeColor:'#2999f0',opacity:0});var t3= b.txt(1.12,2.5," \\textcolor{#02b03e}{\\text{I}}\\text{nner terms}",{"fontsize":1.3,opacity:0}); var f3= b.bendyArrow([0.77-2,0.8],[1.48-2,0.8],-40,-140,{strokeColor:'#02b03e',opacity:0});var t4 = b.txt(1.12,2.5," \\textcolor{#a72cd4}{\\text{L}}\\text{ast terms}",{"fontsize":1.3,opacity:0}); var f4 = b.bendyArrow([0.77-2,0.8],[1.98-2,0.8],-50,-150,{strokeColor:'#a72cd4',opacity:0});var p1 = b.txt(0.85,1,"= \\ ac",{fontSize:1.3,opacity:0}); var l1 = b.txt(1.1,1.65," \\ \\textcolor{red}{\\text{F}}",{fontSize:1.3,opacity:0});var p2 =b.txt(1.75,1,"+ \\ ad",{fontSize:1.3,opacity:0}); var l2 =b.txt(1.9,1.65," \\ \\textcolor{#2999f0}{\\text{O}}",{fontSize:1.3,opacity:0});var p3 = b.txt(2.65,1,"+ \\ bc",{fontSize:1.3,opacity:0}); var l3 = b.txt(2.8,1.65," \\ \\textcolor{#02b03e}{\\text{I}}",{fontSize:1.3,opacity:0});var p4 =b.txt(3.55,1,"+ \\ bd",{fontSize:1.3,opacity:0}); var l4 =b.txt(3.7,1.65," \\ \\textcolor{#a72cd4}{\\text{L}}",{fontSize:1.3,opacity:0});mlg.af("A",function() { $('#btn1').css({'background-color':'#24c5ed'});b.hide(p1,0); b.hide(l1,0); b.hide(f1,0); b.hide(p2,0); b.hide(l2,0); b.hide(f2,0); b.hide(p3,0); b.hide(l3,0); b.hide(f3,0); b.hide(p4,0); b.hide(l4,0); b.hide(f4,0);b.show(f1,t); setTimeout( function() { b.show(t1,t); }, 2*t); setTimeout( function() { b.hide(t1,t); }, 4*t); setTimeout( function() { b.show(l1,t); }, 4*t); setTimeout( function() { b.show(p1,t); }, 4*t);setTimeout( function() { b.show(f2,t); }, 6*t); setTimeout( function() { b.show(t2,t); }, 8*t); setTimeout( function() { b.hide(t2,t); }, 10*t); setTimeout( function() { b.show(l2,t); }, 10*t); setTimeout( function() { b.show(p2,t); }, 10*t);setTimeout( function() { b.show(f3,t); }, 12*t); setTimeout( function() { b.show(t3,t); }, 14*t); setTimeout( function() { b.hide(t3,t); }, 16*t); setTimeout( function() { b.show(l3,t); }, 16*t); setTimeout( function() { b.show(p3,t); }, 16*t);setTimeout( function() { b.show(f4,t); }, 18*t); setTimeout( function() { b.show(t4,t); }, 20*t); setTimeout( function() { b.hide(t4,t); }, 22*t); setTimeout( function() { b.show(l4,t); }, 22*t); setTimeout( function() { b.show(p4,t); }, 22*t);setTimeout( function() { $('#btn1').css({'background-color':'#d8dce3'}); }, 24*t); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution19954_1_2063897220_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution19954_1_2063897220_l", "Solution19954_1_2063897220_p", 1, code); }); } ); } window.JXQtable["Solution19954_1_2063897220_l"] = true;Calculate Productwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#btn1').on('touchstart mousedown', function () { try { mlg.cf("A"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); | |

Exercises 3 We want to find a product. To do so, we will apply the Distributive Property. (x+1)(x+3)Distribute (x+3)x(x+3)+1(x+3)Distribute xx2+3x+1(x+3)Distribute 1x2+3x+x+3Add termsx2+4x+3 | |

Exercises 4 We want to find a product. To do so, we will apply the Distributive Property. (y+6)(y+4)Distribute (y+4)y(y+4)+6(y+4)Distribute yy2+4y+6(y+4)Distribute 6y2+4y+6y+24Add termsy2+10y+24 | |

Exercises 5 We want to find a product. To do so, we will apply the Distributive Property. (z−5)(z+3)Distribute (z+3)z(z+3)−5(z+3)Distribute zz2+3z−5(z+3)Distribute -5z2+3z−5z−15Subtract termz2−2z−15 | |

Exercises 6 We want to find a product. To do so, we will apply the Distributive Property. (a+8)(a−3)Distribute (a−3)a(a−3)+8(a−3)Distribute aa2−3a+8(a−3)Distribute 8a2−3a+8a−24Add termsa2+5a−24 | |

Exercises 7 We want to find a product. To do so, we will apply the Distributive Property. (g−7)(g−2)Distribute (g−2)g(g−2)−7(g−2)Distribute gg2−2g−7(g−2)Distribute -7g2−2g−7g+14Subtract termg2−9g+14 | |

Exercises 8 We want to find a product. To do so, we will apply the Distributive Property. (n−6)(n−4)Distribute (n−4)n(n−4)−6(n−4)Distribute nn2−4n−6(n−4)Distribute -6n2−4n−6n+24Subtract termn2−10n+24 | |

Exercises 9 We want to find a product. To do so, we will apply the Distributive Property. (3m+1)(m+9)Distribute (m+9)3m(m+9)+1(m+9)Distribute 3m3m2+27m+1(m+9)Distribute 13m2+27m+m+9Add terms3m2+28m+9 | |

Exercises 10 | |

Exercises 11 We want to simplify the given product using a table. (x+3)(x+2) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.x3 xx⋅xx⋅3 22⋅x2⋅3 Now we can calculate the products.x3 xx23x 22x6 Finally, we obtain the answer by adding the products and combining like terms. x2+3x+2x+6⇔x2+5x+6 | |

Exercises 12 We want to simplify the given product using a table. (y+10)(y−5) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.y10 yy⋅yy⋅10 -5-5⋅y-5⋅10 Now we can calculate the products.y10 yy210y -5-5y-50 Finally, we obtain the answer by adding the products and combining like terms. y2+10y−5y−50⇔y2+5y−50 | |

Exercises 13 We want to simplify the given product using a table. (h−8)(h−9) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.h-8 hh⋅hh⋅(-8) -9-9⋅h-9⋅(-8) Now we can calculate the products.h-8 hh2-8h -9-9h72 Finally, we obtain the answer by adding the products and combining like terms. h2−8h−9h+72⇔h2−17h+72 | |

Exercises 14 We want to simplify the given product using a table. (c−6)(c−5) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.c-6 cc⋅cc⋅(-6) -5-5⋅c-5⋅(-6) Now we can calculate the products.c-6 cc2-6c -5-5c30 Finally, we obtain the answer by adding the products and combining like terms. c2−6c−5c+30⇔c2−11c+30 | |

Exercises 15 We want to simplify the given product using a table. (3k−1)(4k+9) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.3k-1 4k4k⋅3k4k⋅(-1) 99⋅3k9⋅(-1) Now we can calculate the products.3k-1 4k12k2-4k 927k-9 Finally, we obtain the answer by adding the products and combining like terms. 12k2−4k+27k−9⇔12k2+23k−9 | |

Exercises 16 We want to simplify the given product using a table. (5g+3)(g+8) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.5g3 gg⋅5gg⋅3 88⋅5g8⋅3 Now we can calculate the products.5g3 g5g23g 840g24 Finally, we obtain the answer by adding the products and combining like terms. 5g2+3g+40g+24⇔5g2+43g+24 | |

Exercises 17 We want to simplify the given product using a table. (-3+2j)(4j−7) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.-32j 4j4j⋅(-3)4j⋅2j -7-7⋅(-3)-7⋅2j Now we can calculate the products.-32j 4j-12j8j2 -721-14jFinally, we obtain the answer by adding the products and combining like terms. -12j+8j2+21−14jCommutative Property of Addition8j2−12j−14j+21Subtract term8j2−26j+21 | |

Exercises 18 We want to simplify the given product using a table. (5d−12)(-7+3d) We will use the terms of the first factor as column headers, and the terms of the second factor as row headers. Each of the cells of the table will consist of the product of its headers.5d-12 -7-7⋅5d-7⋅(-12) 3d3d⋅5d3d⋅(-12) Now we can calculate the products.5d-12 -7-35d84 3d15d2-36dFinally, we obtain the answer by adding the products and combining like terms. -35d+84+15d2−36dCommutative Property of Addition15d2−35d−36d+84Subtract term15d2−71d+84 | |

Exercises 19 In this exercise, we are asked to find the error in multiplication of two binomials. Let's start by multiplying the binomials using the table.t-2 tt2-2t 55t-10 From this we can see, if we add 5t+-2t we get t2+3t−10, which is different than the answer in the exercise. Let's have a closer look.Here we can see the error is in the first step where their is no t2 term. It looks like the (t+5) was not distributed properly. Let's do that correctly. | |

Exercises 20 For this exercise, we need to find the error in the multiplication of two binomials. They used the column method. Let's look for the error.From this, we can see the very common error of not bringing down the negative with the 5, resulting in the error in the last line. Let's correct the error. | |

Exercises 21 We want to simplify the given product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (b)(b)+(b)(7)+(3)(b)+(3)(7)Multiplyb2+7b+3b+21Add termsb2+10b+21 | |

Exercises 22 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (w)(w)+(w)(6)+(9)(w)+(9)(6)Multiplyw2+6w+9w+54Add termsw2+15w+54 | |

Exercises 23 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (k)(k)+(k)(-1)+(5)(k)+(5)(-1)Multiplyk2+(k)(-1)+5k+(5)(-1)a(-b)=-a⋅bk2−k+5k−5Add termsk2+4k−5 | |

Exercises 24 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (x)(x)+(x)(8)+(-4)(x)+(-4)(8)Multiplyx2+8x+(-4)(x)+(-4)(8)(-a)b=-abx2+8x−4x−24Subtract termx2+4x−24 | |

Exercises 25 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (q)(q)+(q)(41)+(-43)(q)+(-43)(41)Multiplyq2+41q+(-43)(q)+(-43)(41)(-a)b=-abq2+41q−43q−163Subtract fractionsq2−42q−163ba=b/2a/2q2−21q−163 | |

Exercises 26 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (z)(z)+(z)(-32)+(-35)(z)+(-35)(-32)Multiplyz2+(z)(-32)+(-35)(z)+(-35)(-32)a(-b)=-a⋅bz2−32z+(-35)(z)+(-35)(-32)(-a)b=-abz2−32z−35z+(-35)(-32)-a(-b)=a⋅bz2−32z−35z+910Subtract fractionsz2−37z+910 | |

Exercises 27 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (9)(2)+(9)(-3r)+(-r)(2)+(-r)(-3r)Multiply18+(9)(-3r)+(-r)(2)+(-r)(-3r)a(-b)=-a⋅b18−27r+(-r)(2)+(-r)(-3r)(-a)b=-ab18−27r−2r+(-r)(-3r)-a(-b)=a⋅b18−27r−2r+3r2Subtract term18−29r+3r2Commutative Property of Addition3r2−29r+18 | |

Exercises 28 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (8)(2x)+(8)(6)+(-4x)(2x)+(-4x)(6)Multiply16x+48+(-4x)(2x)+(-4x)(6)(-a)b=-ab16x+48−8x2−24xCommutative Property of Addition-8x2+16x−24x+48Subtract term-8x2−8x+48 | |

Exercises 29 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (w)(w2)+(w)(3w)+(5)(w2)+(5)(3w)Multiplyw3+3w2+5w2+15wAdd termsw3+8w2+15w | |

Exercises 30 We want to find the product using the FOIL method. Recall that the acronym FOIL stands for First, Outer, Inner, and Last.Let's now simplify the expression by multiplying and combining like terms. (v)(v2)+(v)(8v)+(-3)(v2)+(-3)(8v)Multiplyv3+8v2+(-3)(v2)+(-3)(8v)(-a)b=-abv3+8v2−3v2−24vSubtract termv3+5v2−24v | |

Exercises 31 In this exercise, we combine our skills of multiplying binomials with finding the area of a rectangle. Notice we are given the length and width of the rectangle.Let's substitute the length and width from the rectangle's picture into the equation for area. A=ℓ⋅wℓ=2x−9, w=x+5A=(2x+9)(x+5) Multiply parentheses Distribute x+5A=2x(x+5)+9(x+5)Distribute 2xA=2x2+10x+9(x+5)Distribute 9A=2x2+10x+9x+45Add terms A=2x2+19x+45 | |

Exercises 32 In this exercise, we combine our skills of multiplying binomials with finding the area of a triangle. Notice that we are given the base and height of the rectangle.Let's substitute the base and height from the triangle's picture into the equation for triangle area. A=21⋅b⋅hb=p+1, h=2p−6A=21(p+1)(2p−6) Multiply parentheses Distribute 2p−cA=21[p(2p−6)+1(2p−6)]Distribute pA=21[2p2−6p+1(2p−6)]Distribute 1A=21[2p2−6p+2p−6]Add terms A=21[2p2−4p−6]Distribute 21A=p2−2p−3 | |

Exercises 33 For this exercise, we need to find the area of the shaded region, which is part of a rectangle. In this case, if we look at the picture we can see that the white triangle in the center is half the area of the the rectangle.We can then say that the shaded region is the other half of the rectangle. Let's first find the area of the rectangle. A=ℓ⋅wℓ=x+6, w=x+5A=(x+6)(x+5) Multiply parentheses Distribute x+5A=x(x+5)+6(x+5)Distribute xA=x2+5x+6(x+5)Distribute 6A=x2+5x+6x+30Add terms A=x2+11x+30 Now, let's take half of the rectangle area to find the polynomial that represents the shaded region. 2x2+11x+30=0.5x2+5.5x+15 | |

Exercises 34 In this exercise, we combine our skills of multiplying binomials with finding the area of a shaded region. This shaded region is a square with a rectangle cut out. We need to find the area of both, then subtract.Let's use the lengths of the sides in the image to find the area of the white rectangle. AAA===ℓ⇓(x−7)⇓(5x−35)⋅⋅w⇓5 When we distribute, we get the area of the white rectangle to be 5x−35. Now, let's find the area of the large square. A=ℓ⋅wℓ=x+1, w=x+1A=(x+1)(x+1) Multiply parentheses Distribute x+1A=x(x+1)+1(x+1)Distribute xA=x2+x+1(x+1)Distribute 1A=x2+x+x+1Add terms A=x2+2x+1 Now, we can subtract the area of the white rectangle from the area of the large square and we will have only the shaded area remaining. A(Shaded Region)↓A(Shaded Region)=↓=A(Large Square)↓(x2+2x+1)−↓−A(White Rectangle)↓(5x−35) Now, let's simplify that expression by combining like terms A=x2+2x+1−(5x−35) Simplify right-hand side Distribute -1A=x2+2x+1−5x+35Commutative Property of AdditionA=x2+2x−5x+1+35Subtract termsA=x2−3x+1+35Add terms A=x2−3x+36 | |

Exercises 35 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. x2+3xx++24 Now, let's multiply x2+3x+2 by 4. x24x2++3xx12x+++248 Now, we will multiply x2+3x+2 by x. x3+x24x23x2+++3xx12x2x+++248 Finally, let's add the like terms in 4x2+12x+8 and x3+3x2+2x. x3x3++x24x23x27x2++++3xx12x2x14x++++2488 The product is x3+7x2+14x+8. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (x+4)(x2+3x+2)Distribute (x2+3x+2)x(x2+3x+2)+4(x2+3x+2)Distribute xx3+3x2+2x+4(x2+3x+2)Distribute 4x3+3x2+2x+4x2+12x+8Commutative Property of Additionx3+3x2+4x2+2x+12x+8Add termsx3+7x2+14x+8 Both methods are correct and give the same result. | |

Exercises 36 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. f2+4ff++81 Now, let's multiply f2+4f+8 by 1. f2f2++4ff4f+++818 Now, we will multiply f2+4f+8 by f. f3+f2f24f2+++4ff4f8f+++818 Finally, let's add the like terms in f2+4f+8 and f3+4f2+8f. f3f3++f2f24f25f2++++4ff4f8f12f++++8188 The product is f3+5f2+12f+8. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (f+1)(f2+4f+8)Distribute (f2+4f+8)f(f2+4f+8)+1(f2+4f+8)Distribute ff3+4f2+8f+1(f2+4f+8)Distribute 1f3+4f2+8f+f2+4f+8Commutative Property of Additionf3+4f2+f2+8f+4f+8Add termsf3+5f2+12f+8 Both methods are correct and give the same result. | |

Exercises 37 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. y2+8yy−+23 Now, let's multiply y2+8y−2 by 3. y23y2++8yy24y−+−236 Now, we will multiply y2+8y−2 by y. y3+y23y28y2++−8yy24y2y−+−236 Finally, let's add the like terms in 3y2+24y−6 and y3+8y2−2y. y3y3++y23y28y211y2++−+8yy24y2y22y−+−−2366 The product is y3+11y2+22y−6. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (y+3)(y2+8y−2)Distribute (y2+8y−2)y(y2+8y−2)+3(y2+8y−2)Distribute yy3+8y2−2y+3(y2+8y−2)Distribute 3y3+8y2−2y+3y2+24y−6Commutative Property of Additiony3+8y2+3y2−2y+24y−6Add termsy3+11y2+22y−6 Both methods are correct and give the same result. | |

Exercises 38 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. t2−5tt+−12 Now, let's multiply t2−5t+1 by -2. t2-2t2−+5tt10t+−−122 Now, we will multiply t2−5t+1 by t. t3−t2-2t25t2−++5tt10tt+−−122 Finally, let's add the like terms in -2t2+10t−2 and t3−5t2+t. t3t3−−t2-2t25t27t2−+++5tt10tt11t+−−−1222 The product is t3−7t2+11t−2. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (t−2)(t2−5t+1)Distribute (t2−5t+1)t(t2−5t+1)−2(t2−5t+1)Distribute tt3−5t2+t−2(t2−5t+1)Distribute -2t3−5t2+t−2t2+10t−2Commutative Property of Additiont3−5t2−2t2+t+10t−2Add and subtract termst3−7t2+11t−2 Both methods are correct and give the same result. | |

Exercises 39 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 5b2+5b-b−+44 Now, let's multiply 5b2+5b−4 by 4. 5b220b2++5b-b20b−+−4416 Now, we will multiply 5b2+5b−4 by -b. -5b3−5b220b25b2+++5b-b20b4b−+−4416 Finally, let's add 20b2+5b−16 and -5b3−5b2+4b like terms. -5b3-5b3−+5b220b25b215b2++++5b-b20b4b24b−+−−441616 The product is -5b3+15b2+24b−16. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (4−b)(5b2+5b−4)Distribute (5b2+5b−4)4(5b2+5b−4)−b(5b2+5b−4)Distribute 420b2+20b−16−b(5b2+5b−4)Distribute -b20b2+20b−16−5b3−5b2+4bCommutative Property of Addition-5b3+20b2−5b2+20b+4b−16Add and subtract terms-5b3+15b2+24b−16 Both method are correct and gives the same result. | |

Exercises 40 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 2d2−dd++76 Now, let's multiply 2d2−d+7 by 6. 2d212d2−−dd6d+++7642 Now, we will multiply 2d2−d+7 by d. 2d3−2d212d2d2−−+dd6d7d+++7642 Finally, let's add the like terms in 12d2−6d+42 and 2d3−d2+7d. 2d32d3−+2d212d2d211d2−−++dd6d7dd++++764242 The product is 2d3+11d2+d+42. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (d+6)(2d2−d+7)Distribute (2d2−d+7)d(2d2−d+7)+6(2d2−d+7)Distribute d2d3−d2+7d+6(2d2−d+7)Distribute 62d3−d2+7d+12d2−6d+42Commutative Property of Addition2d3−d2+12d2+7d−6d+42Add and subtract terms2d3+11d2+d+42 Both methods are correct and give the same result. | |

Exercises 41 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 3e2−5e6e++71 Now, let's multiply 3e2−5e+7 by 1. 3e23e2−−5e6e5e+++717 Now, we will multiply 3e2−5e+7 by 6e. 18e3−3e23e230e2−−+5e6e5e42e+++717 Finally, let's add the like terms in 3e2−5e+7 and 18e3−30e2+42e. 18e318e3−−3e23e230e227e2−−++5e6e5e42e37e++++7177 The product is 18e3−27e2+37e+7. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (6e+1)(3e2−5e+7)Distribute (3e2−5e+7)6e(3e2−5e+7)+1(3e2−5e+7)Distribute 6e18e3−30e2+42e+1(3e2−5e+7)Distribute 118e3−30e2+42e+3e2−5e+7Commutative Property of Addition18e3−30e2+3e2+42e−5e+7Add and subtract terms18e3−27e2+37e+7 Both methods are correct and give the same result. | |

Exercises 42 To simplify the given expression, we will multiply a binomial and a trinomial. Let's align the like terms vertically. 6v2+2v-5v−+94 Now, let's multiply 6v2+2v−9 by 4. 6v224v2++2v-5v8v−+−9436 Now, we will multiply 6v2+2v−9 by -5v. -30v3−6v224v210v2+++2v-5v8v45v−+−9436 Finally, let's add the like terms in 24v2+8v−36 and -30v3−10v2+45v. -30v3-30v3−+6v224v210v214v2++++2v-5v8v45v53v−+−−943636 The product is -30v3+14v2+53v−36. Alternative solution info Simplifying by Distribution We can also simplify the given expression by applying the Distributive Property. (6v2+2v−9)(4−5v)Distribute (4−5v)6v2(4−5v)+2v(4−5v)−9(4−5v)Distribute 6v224v2−30v3+2v(4−5v)−9(4−5v)Distribute 2v24v2−30v3+8v−10v2−9(4−5v)Distribute -924v2−30v3+8v−10v2−36+45vCommutative Property of Addition-30v3+24v2−10v2+8v+45v−36Add and subtract terms-30v3+14v2+53v−36 Both methods are correct and give the same result. | |

Exercises 43 | |

Exercises 44 | |

Exercises 45 Let's start by recalling the FOIL Method for multiplying binomials. The word FOIL is an acronym for the words First, Outer, Inner, and Last. This is a mnemonic to remind us the order to follow when multiplying binomials.Let's see an example. Consider that we want to find the product of the binomials (x2+2) and (3x3+5). Since the degree of a polynomial is determined by the highest degree of its monomials, we know we are multiplying a degree 2 and a degree 3 binomial.As we can see, the product of a degree 2 polynomial and a degree 3 binomial results in a degree 5 polynomial. This happened because when the terms determining the degree of each binomial were multiplied, the result was a monomial of a degree equal to the sum of the degree of each individual monomial factor. (x2+2)(3x3+5)=3x5+5x2+6x3+10x2⋅3x3=3x2+3=3x5 The same will happen with any pair of binomials in general. Therefore, we can summarize the ideas we have discussed as shown below.The degree of the product of two binomials is equal to the sum of the degrees of each binomial factor. | |

Exercises 46 We are asked for a pair of polynomials such that their product is a trinomial of degree 3. For simplicity, we will look for two binomials with the required characteristics. Recall that we can find the product of two binomials using the FOIL Method, which is an acronym for the words First, Outer, Inner, and Last.Furthermore, recall that the degree of the product of two binomials is equal to the sum of the degrees of each binomial. We can try with a degree 1 binomial and a degree 2 binomial. For example, (x+2) and (x2+5). Lets multiply them using the FOIL Method.As we can see, we obtained a degree 3 polynomial as result. However, since none of the products are like terms, we cannot simplify it to obtain a trinomial. Nevertheless, if we use (x2+5x) instead of (x2+5), two of the product terms are quadratic and they can be combined, obtaining a trinomial as required.The same would happen with any product of the form (ax+b)(cx2+dx). Therefore, there are infinitely many binomials satisfying the exercise's requirements, and this is just an example. | |

Exercises 47 Recall that the word FOIL is an acronym for the words First, Outer, Inner, and Last. This is a mnemonic to remind us of the order to follow when multiplying two binomials.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution19999_1_698584143_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b=mlg.board([-2.38,3,4.62,0],{"grid":{"visible":false},"desktopSize":"large", "style":"usa"}); var t =600;b.txt(0-2,1,"(",{fontSize:1.15}); b.txt(0.25-2,1,"a",{fontSize:1.15}); b.txt(0.5-2,1,"+",{fontSize:1.15}); b.txt(0.75-2,1,"b",{fontSize:1.15}); b.txt(1.0-2,1,")",{fontSize:1.15}); b.txt(1.25-2,1,"(",{fontSize:1.15}); b.txt(1.5-2,1,"c",{fontSize:1.15}); b.txt(1.75-2,1,"+",{fontSize:1.15}); b.txt(2.0-2,1,"d",{fontSize:1.15}); b.txt(2.25-2,1,")",{fontSize:1.15});var t1 = b.txt(1.12,2.5," \\textcolor{red}{\\text{F}}\\text{irst terms}",{"fontsize":1.15,opacity:0}); var f1 = b.bendyArrow([0.27-2,1.20],[1.48-2,1.2],40,140,{strokeColor:'red',opacity:0});var t2= b.txt(1.12,2.5," \\textcolor{#2999f0}{\\text{O}}\\text{uter terms}",{"fontsize":1.15,opacity:0}); var f2= b.bendyArrow([0.27-2,1.20],[1.98-2,1.2],50,130,{strokeColor:'#2999f0',opacity:0});var t3= b.txt(1.12,2.5," \\textcolor{#02b03e}{\\text{I}}\\text{nner terms}",{"fontsize":1.15,opacity:0}); var f3= b.bendyArrow([0.77-2,0.8],[1.48-2,0.8],-40,-140,{strokeColor:'#02b03e',opacity:0});var t4 = b.txt(1.12,2.5," \\textcolor{#a72cd4}{\\text{L}}\\text{ast terms}",{"fontsize":1.15,opacity:0}); var f4 = b.bendyArrow([0.77-2,0.8],[1.98-2,0.8],-50,-150,{strokeColor:'#a72cd4',opacity:0});var p1 = b.txt(0.85,1,"= \\ ab",{fontSize:1.15,opacity:0}); var l1 = b.txt(1.1,1.65," \\ \\textcolor{red}{\\text{F}}",{fontSize:1.15,opacity:0});var p2 =b.txt(1.75,1,"+ \\ ad",{fontSize:1.15,opacity:0}); var l2 =b.txt(1.9,1.65," \\ \\textcolor{#2999f0}{\\text{O}}",{fontSize:1.15,opacity:0});var p3 = b.txt(2.65,1,"+ \\ bc",{fontSize:1.15,opacity:0}); var l3 = b.txt(2.8,1.65," \\ \\textcolor{#02b03e}{\\text{I}}",{fontSize:1.15,opacity:0});var p4 =b.txt(3.55,1,"+ \\ bd",{fontSize:1.15,opacity:0}); var l4 =b.txt(3.7,1.65," \\ \\textcolor{#a72cd4}{\\text{L}}",{fontSize:1.15,opacity:0});mlg.af("A",function() { $('#btn1').css({'background-color':'#24c5ed'});b.hide(p1,0); b.hide(l1,0); b.hide(f1,0); b.hide(p2,0); b.hide(l2,0); b.hide(f2,0); b.hide(p3,0); b.hide(l3,0); b.hide(f3,0); b.hide(p4,0); b.hide(l4,0); b.hide(f4,0);b.show(f1,t); setTimeout( function() { b.show(t1,t); }, 2*t); setTimeout( function() { b.hide(t1,t); }, 4*t); setTimeout( function() { b.show(l1,t); }, 4*t); setTimeout( function() { b.show(p1,t); }, 4*t);setTimeout( function() { b.show(f2,t); }, 6*t); setTimeout( function() { b.show(t2,t); }, 8*t); setTimeout( function() { b.hide(t2,t); }, 10*t); setTimeout( function() { b.show(l2,t); }, 10*t); setTimeout( function() { b.show(p2,t); }, 10*t);setTimeout( function() { b.show(f3,t); }, 12*t); setTimeout( function() { b.show(t3,t); }, 14*t); setTimeout( function() { b.hide(t3,t); }, 16*t); setTimeout( function() { b.show(l3,t); }, 16*t); setTimeout( function() { b.show(p3,t); }, 16*t);setTimeout( function() { b.show(f4,t); }, 18*t); setTimeout( function() { b.show(t4,t); }, 20*t); setTimeout( function() { b.hide(t4,t); }, 22*t); setTimeout( function() { b.show(l4,t); }, 22*t); setTimeout( function() { b.show(p4,t); }, 22*t);setTimeout( function() { $('#btn1').css({'background-color':'#d8dce3'}); }, 24*t); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution19999_1_698584143_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution19999_1_698584143_l", "Solution19999_1_698584143_p", 1, code); }); } ); } window.JXQtable["Solution19999_1_698584143_l"] = true;Calculate Productwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#btn1').on('touchstart mousedown', function () { try { mlg.cf("A"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); Therefore, the FOIL Method can only be used for multiplying two binomials. When two trinomials are multiplied there are 9 products as result. Below we can see the result of multiplying two general trinomials, (a+b+c), and (d+e+f).Hence, if we follow the FOIL Method to multiply two trinomials we would be missing 5 terms. | |

Exercises 48 | |

Exercises 49 In this exercise, we need to show that the two methods for multiplying polynomials results in the same solution. Let's start by showing each method separately, then comparing the parts.Distributive Property Method Let's show how to multiply two binomials using the distributive property. (x+2)(x−5)Distribute x−5x(x−5)+2(x−5)Distribute xx2−5x+2(x−5)Distribute 2x2−5x+2x−10Add termsx2−3x−10FOIL Method Now, Let's show what happens when we use FOIL Method. Let's look at each piece.Fx2 O2x I-5x L-10 This gives us the exact same pieces as the Distributive Property. Furthermore, if we add the linear terms above, we get the same answer as the previous method, x2−3x−10 even though the factors were swapped. (x+2)(x−5)=(x−5)(x+2) The factors can be swapped because multiplication is commutative. | |

Exercises 50 For this exercise, we need to use the formula for the volume of a rectangular prism to get the polynomial for the volume of the container. Let's substitute ℓ=4x−3 and w=x+1 to find the area of the rectangular base. Then we can multiply the answer by h=x+2 to find the volume. A=ℓ⋅wℓ=4x−3, w=x+1A=(4x−3)(x+1) Multiply parentheses Distribute x+1A=4x(x+1)−3(x+1)Distribute 4xA=4x2+4x−3(x+1)Distribute -3A=4x2+4x−3x−3Subtract terms A=4x2+x−3 Now we have the area of the base we can multiply it by our height to find our volume. Let's use the table method to multiply (x+2)(4x2+x−3).×4x2x-3 x4x3x2-3x 28x22x-6 Now, we can simplify by adding like terms to get the polynomial for the area of the shaded region, 4x3+9x2−x−6. | |

Exercises 51 | |

Exercises 52 | |

Exercises 53 | |

Exercises 54 | |

Exercises 55 | |

Exercises 56 | |

Exercises 57 | |

Exercises 58 |

##### Other subchapters in Polynomial Equations and Factoring

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Adding and Subtracting Polynomials
- Special Products of Polynomials
- Solving Polynomial Equations in Factored Form
- Quiz
- Factoring x² + bx + c
- Factoring ax² + bx + c
- Factoring Special Products
- Factoring Polynomials Completely
- Chapter Review
- Chapter Test
- Cumulative Assessment