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Exercises 1 Recall that the degree of a polynomial is the highest degree of the monomials composing it. For example, the polonomial shown below is of degree 4. 2x−5+3x3−x4+9x2 Therefore, to determine the degree of a polynomial it is convenient to have it written in order, such that the exponents of its terms decrease from left to right. When a polynomial is written this way we say it is in standard form.Example Polynomial 2x−5+3x3−x4+9x2 Polynomial in Standard Form -x4+3x3+9x2+2x−5 | |

Exercises 2 Let's start by recalling the definition of a trinomial. A trinomial is a polynomial of three terms, and therefore it follows the same rules. Its degree, just as the degree of any polynomial, is the highest degree of the monomials that form it. For example, the polynomial shown below is of degree 4. 2x−5+3x3−x4+9x2 When a polynomial is written in order, such that the exponents of its terms decrease from left to right, we say it is in standard form.Example Polynomial 2x−5+3x3−x4+9x2 Polynomial in Standard Form -x4+3x3+9x2+2x−5 Therefore, to write a trinomial of degree 5 in standard form, we need three things.A polynomial with three terms The terms must be ordered so that the exponents from its terms are decreasing from left to right The highest degree of the monomials that form it must be 5 An example fulfilling these requirements is shown below. 4x5−3x2+8 Notice that there are infinitely many trinomials satisfying the exercise's conditions. This is just one example. | |

Exercises 3 Let's start by reviewing the concept of closure. We say that a set of numbers is closed under an operation when the operation is performed on any two numbers in the set, and results in a number that is also in the set. For example, consider the set of whole numbers and the operation of subtraction.Set of numbersOperation Whole numbers {0, 1, 2, 3, 4, …}Subtraction ‘‘−"Note that 2 and 3 belong to the set of whole numbers. {0, 1, 2, 3, 4, …} Now, let's apply the operation we are testing — in this case, subtraction. 2−3=-1 Since -1 is not an element of the set of whole numbers, by the definition of closure the set of whole numbers is not closed under subtraction. Note that sometimes it is easier to show that a set is not closed under an operation, since we can prove it by counterexample. We can summarize all these as idea as shown below.To determine if the whether a set of numbers is closed under an operation, we need to determine if the operation performed on any two numbers in the set results in a number that is also in the set. | |

Exercises 4 We are given the expressions shown below, and we are asked to find the one that is different. a3+4ax2−8xb−2-1-3π+6y8z Note that these expressions seem to be polynomials, so let's can check which of them are. Recall that every term of a polynomial must be a monomial. A monomial is a number, a variable, or the product of a number and one or more variables with whole number exponents. Let's have a second look to the expressions given. a3+4ax2−8xb−2-1-3π+6y8z Note that all the expressions satisfy the conditions except for x2−8x, which does not have a whole number as exponent, but a variable instead. Therefore, this is the different one. It might seem that the expression b−2-1 has the same problem, but it does not. We can rewrite it to make this explicit. b−2-1⇔b1−2-1b0 Recall that a nonzero constant is a term of degree zero. As we can see, in this expression it is not the variable, but the coefficient has a negative exponent. Since this is not against the definition, this expression is a polynomial. | |

Exercises 5 A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. The degree of a monomial is the sum of the exponents of its variables. The degree of a nonzero constant is 0. 4g⇔4g1 The degrees of g is 1. So the degree of the monomial is 1. | |

Exercises 6 A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. The degree of a monomial is the sum of the exponents of its variables. The degree of a nonzero constant is 0. 23x4 The degrees of x is 4. So the dergree of the monomial is 4. | |

Exercises 7 A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. The degree of a monomial is the sum of the exponents of its variables. The degree of a nonzero constant is 0. -1.75k2 The degrees of k is 2. So the dergree of the monomial is 2. | |

Exercises 8 A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. The degree of a monomial is the sum of the exponents of its variables. The degree of a nonzero constant is 0. -94 -94 is a nonzero constant, so the degree of the monomial is 0. | |

Exercises 9 A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. The degree of a monomial is the sum of the exponents of its variables. The degree of a nonzero constant is 0. s2t⇔s2t1 The degrees of s and t are 2 and 1, respectively. We find the degree of the monomial by adding 2 and 1. 2+1=3 | |

Exercises 10 A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. The degree of a monomial is the sum of the exponents of its variables. The degree of a nonzero constant is 0. 8m2n4 The degrees of m and n are 2 and 4, respectively. We find the degree of the monomial by adding 2 and 4. 2+4=6 | |

Exercises 11 A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. The degree of a monomial is the sum of the exponents of its variables. The degree of a nonzero constant is 0. 9xy3z7⇔9x1y3z7 The degrees of x, y and z are 1, 3 and 7, respectively. We find the degree of the monomial by adding 1, 3 and 7. 1+3+7=11 | |

Exercises 12 The degree of a monomial is the sum of the exponents of the variables. In this monomial, our variables are q,r, and s. -3q4rs6⇒-3q4r1s6 We can see that the exponents of the variables are 4,1, and 6. To find the degree, we will add these values together. 4+1+6=11 The degree of the monomial is 11. | |

Exercises 13 The standard form of a polynomial arranges the terms by degree in descending numerical order. 6c2+2c4−c⇔2c4+6c2−c The degree of a polynomial is the highest exponent of the variable. 2c4+6c2−c The highest exponent of the polynomial, and therefore its degree, is 4. The leading coefficient of the polynomial is 2. We want to classify this polynomial by the number of terms. This polynomial consists of three terms, so it is a trinomial. | |

Exercises 14 The standard form of a polynomial arranges the terms by degree in descending numerical order. 4w11−w12⇔-w12+4w11 The degree of a polynomial is the highest exponent of the variable. -1w12+4w11 The highest exponent of the polynomial, and therefore its degree, is 12. The leading coefficient of the polynomial is -1. We want to classify this polynomial by the number of terms. This polynomial consists of two terms, so it is a binomial. | |

Exercises 15 The standard form of a polynomial arranges the terms by degree in descending numerical order. 7+3p2⇔3p2+7 The degree of a polynomial is the highest exponent of the variable. 3p2+7 The highest exponent of the polynomial, and therefore its degree, is 2. The leading coefficient of the polynomial is 3. We want to classify this polynomial by the number of terms. This polynomial consists of two terms, so it is a binomial. | |

Exercises 16 The standard form of a polynomial arranges the terms by degree in descending numerical order. 8d−2−4d3⇕-4d3+8d−2 We want to classify this polynomial by degree, leading coefficient, and the number of terms. The degree of a polynomial is the highest exponent of the variable. Let's identify leading coefficient and degree. -4d3+8d−2 The highest exponent of the polynomial, and therefore its degree, is 3. The leading coefficient of the polynomial is -4. This polynomial consists of three terms, so it is a trinomial. | |

Exercises 17 The standard form of a polynomial arranges the terms by degree in descending numerical order. 3t8 Since we have only one term, our expression is already in standard form. The degree of a polynomial is the highest exponent of the variable. 3t8 The highest exponent of the polynomial, and therefore its degree, is 8. The leading coefficient of the polynomial is 3. We want to classify this polynomial the number of terms. This polynomial consists of one term, so it is a monomial. | |

Exercises 18 The standard form of a polynomial arranges the terms by degree in descending numerical order. 5z+2z3+3z4⇕3z4+2z3+5z We want to classify this polynomial by degree, leading coefficient, and the number of terms. The degree of a polynomial is the highest exponent of the variable. Let's identify leading coefficient and degree. 3z4+2z3+5z The highest exponent of the polynomial, and therefore its degree, is 4. The leading coefficient of the polynomial is 3. This polynomial consists of three terms, so it is a trinomial. | |

Exercises 19 The standard form of a polynomial arranges the terms by degree in descending numerical order. πr2−75r8+2r5⇔-75r8+2r5+πr2 The degree of a polynomial is the highest exponent of the variable. -75r8+2r5+πr2 The highest exponent of the polynomial, and therefore its degree, is 8. The leading coefficient of the polynomial is -75. We want to classify this polynomial by the number of terms. This polynomial consists of three terms, so it is a trinomial. | |

Exercises 20 The standard form of a polynomial arranges the terms by degree in descending numerical order. 7n4 Since we have only one term, our expression is already in standard form. The degree of a polynomial is the highest exponent of the variable. 7n4 The highest exponent of the polynomial, and therefore its degree, is 4. The leading coefficient of the polynomial is 7. We want to classify this polynomial the number of terms. This polynomial consists of one term, so it is a monomial. | |

Exercises 21 Showing it is a Monomial We want to describe why 34πr3 is a monomial. The definition of a monomial tells us that a monomial can be the product of a number and variables raised to whole number exponents. 34π⋅r3≈4.19⋅r3 Notice that our expression is a number multiplied by a variable with an exponent. Furthermore, remember that π is just a number, and is not a variable. Therefore, the expression fits the definition of a monomial.Finding the Degree The degree of a monomial is the sum of the exponents of the variables. In this monomial, our only variable is r and it is raised to the third power. 34πr3 Since r is our only variable, the degree of this monomial is 3. | |

Exercises 22 We will start by classifying the polynomial 400x3+600x6 by its number of terms. A term is a monomial. Notice that this polynomial is the addition of two monomials. 400x3+600x6 Therefore, it has two terms. The name for a polynomial with two terms is a binomial. Now, we need to find the degree of this polynomial. The degree of a polynomial is defined as the greatest degree of its terms. 400x3+600x6 The greatest degree of the terms is 6, so 6 is the degree of the polynomial. | |

Exercises 23 The first step in finding a sum is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (5y+4)+(-2y+6) In this case, we have two y-terms, and two constants. Both the y-terms and constants can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (5y+4)+(-2y+6)Remove parentheses5y+4−2y+6Commutative Property of Addition5y−2y+4+6Subtract term3y+4+6Add terms3y+10 | |

Exercises 24 The first step in finding a sum is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (-8x−12)+(9x+4) In this case, we have two x-terms, and two constants. Both x-terms and constants can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (-8x−12)+(9x+4)Remove parentheses-8x−12+9x+4Commutative Property of Addition-8x+9x−12+4Add termsx−8 | |

Exercises 25 The first step in simplifying this expression is to identify which, if any, terms can be combined. Remember that only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (2n2−5n−6)+(-n2−3n+11) In this case, we have three pairs of like terms, two n2-terms, two n-terms, and two constants. To simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (2n2−5n−6)+(-n2−3n+11)Remove parentheses2n2−5n−6−n2−3n+11Commutative Property of Addition2n2−n2−5n−3n−6+11Add and subtract termsn2−8n+5 | |

Exercises 26 The first step in finding a sum is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (-3p3+5p2−2p)+(-p3−8p2−15p) In this case, we have two p3-terms, two p2-terms and two p-terms. All terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (-3p3+5p2−2p)+(-p3−8p2−15p)Remove parentheses-3p3+5p2−2p−p3−8p2−15pCommutative Property of Addition-3p3−p3+5p2−8p2−2p−15pSubtract terms-4p3−3p2−17p | |

Exercises 27 The first step in finding a sum is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (3g2−g)+(3g2−8g+4) In this case, we have two g2-terms, two g-term, and one constant. Only the g2-terms and g-terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (3g2−g)+(3g2−8g+4)Remove parentheses3g2−g+3g2−8g+4Commutative Property of Addition3g2+3g2−g−8g+4Add terms6g2−g−8g+4Subtract term6g2−9g+4 | |

Exercises 28 The first step in finding a sum is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (9r2+4r−7)+(3r2−3r) In this case, we have two r2-terms, two r-term, and one constant. Only the r2-terms and r-terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (9r2+4r−7)+(3r2−3r)Remove parentheses9r2+4r−7+3r2−3rCommutative Property of Addition9r2+3r2+4r−3r−7Add terms6r2+4r−3r−7Subtract term6r2+r−7 | |

Exercises 29 The first step in finding a sum is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (4a−a3−3)+(2a3−5a2+8) In this case, we have two a3-terms, one a-term, one a-term and two constants. Only the a3-terms and constant-terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (4a−a3−3)+(2a3−5a2+8)Remove parentheses4a−a3−3+2a3−5a2+8Commutative Property of Addition-a3+2a3−5a2+4a−3+8Add termsa3−5a2+4a+5 | |

Exercises 30 The first step in finding a sum is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (s3−2s−9)+(2s2−6s3+s) In this case, we have two s3-terms, one s2-term, two s-terms and one constant. Only the s3-terms and s-terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (s3−2s−9)+(2s2−6s3+s)Remove parenthesess3−2s−9+2s2−6s3+sCommutative Property of Additions3−6s3+2s2−2s+s−9Subtract term-5s3+2s2−2s+s−9Add terms-5s3+2s2−s−9 | |

Exercises 31 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (d−9)−(3d−1) In this case, we have two d-terms, and two constants. All terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (d−9)−(3d−1)Distribute -1(d−9)−3d+1Remove parenthesesd−9−3d+1Commutative Property of Additiond−3d−9+1Subtract term-2d−9+1Add terms-2d−8 | |

Exercises 32 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (6x+9)−(7x+1) In this case, we have two x-terms, and two constants. All terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (6x+9)−(7x+1)Distribute -1(6x+9)−7x−1Remove parentheses6x+9−7x−1Commutative Property of Addition6x−7x+9−1Subtract term-x+8 | |

Exercises 33 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (y2−4y+9)−(3y2−6y−9) In this case, we have two y2-terms, two y-terms and two constants. All terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (y2−4y+9)−(3y2−6y−9)Distribute -1(y2−4y+9)−3y2+6y+9Remove parenthesesy2−4y+9−3y2+6y+9Commutative Property of Additiony2−3y2−4y+6y+9+9Subtract term-2y2−4y+6y+9+9Add terms-2y2+2y+18 | |

Exercises 34 To find the difference, we will use the Distributive Property to eliminate the second set of parentheses. (4m2−m+2)−(-3m2+10m+4)Remove parentheses4m2−m+2−(-3m2+10m+4)Distribute -14m2−m+2+3m2−10m−4 The next step in simplifying this expression is to identify which, if any, terms can be combined. Remember that only like terms — constant terms or terms with the same variable and the same exponent — can be combined. 4m2−m+2+3m2−10m−4 In this case, we have three pairs of like terms, two m2-terms, two m-terms, and two constants. To simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. 4m2−m+2+3m2−10m−4Commutative Property of Addition4m2+3m2−m−10m+2−4Add and subtract terms7m2−11m−2 | |

Exercises 35 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (k3−7k+2)−(k2−12) In this case, we have one k3-term, one k2-ter, one k-term and two constants. Only constants can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (k3−7k+2)−(k2−12)Distribute -1(k3−7k+2)−k2+12Remove parenthesesk3−7k+2−k2+12Commutative Property of Additionk3−k2−7k+2+12Add termsk3−k2−7k+14 | |

Exercises 36 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (-r−10)−(-4r3+r2+7r) In this case, we have one r3-term, one r2-term, two r-terms and one constant. Only r-terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (-r−10)−(-r3+r2+7r)Distribute -1(-r−10)+r3−r2−7rRemove parentheses-r−10+r3−r2−7rCommutative Property of Additionr3−r2−r−7r−10Subtract termr3−r2−8r−10 | |

Exercises 37 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (t4−t2+t)−(12−9t2−7t) In this case, we have one t4-term, two t2-terms, two t-terms and one constant. Only t2-terms and t-terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (t4−t2+t)−(12−9t2−7t)Distribute -1(t4−t2+t)−12+9t2+7tRemove parenthesest4−t2+t−12+9t2+7tCommutative Property of Additiont4−t2+9t2+t+7t−12Add termst4+8t2+8t−12 | |

Exercises 38 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (4d−6d3+3d2)−(10d3+7d+2) In this case, we have two d3-terms, one d2-term, two d-terms and one constant. Only d3-terms and d-terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (4d−6d3+3d2)−(10d3+7d+2)Distribute -1(4d−6d3+3d2)−10d3−7d−2Remove parentheses4d−6d3+3d2−10d3−7d−2Commutative Property of Addition-6d3−10d3+3d2+4d−7d−2Subtract terms-16d3+3d2−3d−2 | |

Exercises 39 Finding the Error We will start by finding the error that was made. Let's look at each step. (x2+x)−(2x2−3x)=x2+x−2x2−3x× The error was made in this first step — there was an error in distributing the subtraction sign to the second binomial. The negative sign should be distributed to each term in the second binomial. Recall that two negatives make a positive. -(-a)=aCorrectly Solving the Difference We will now correctly solve this problem. (x2+x)−(2x2−3x)1⋅a=a(x2+x)−1(2x2−3x)Distribute -1x2+x−2x2−(-3x)a−(-b)=a+bx2+x−2x2+3xCommutative Property of Addition(x2−2x2)+(x+3x)Add and subtract terms-x2+4x When correctly simplified, the original expression equals -x2+4x. | |

Exercises 40 Finding the Error We will start by finding the error that was made. x3+ -3x3-2x3−4x2+3+8x − 2+4x2+1× The error was made in adding these two expressions. Remember that when we add polynomials we can only add like terms. This means we can only add x2 terms with other x2 terms. Similarly, we can only add x terms with x terms. Notice that in the work shown, -4x2 is added to 8x. This is the error. -+ 4x28x4x2×Correctly Solving the Difference We will now correctly solve this problem. To make sure we are adding like terms, let's start by aligning the like terms vertically. x3+ -3x3−4x2 +3 +8x − 2 Now we can add each like term. x3+ -3x3-2x3−4x2 +3 +8x −2−4x2+8x +1 When we correctly add these polynomials, we get -2x3−4x2+8x+1. | |

Exercises 41 We want to write an expression that represents how much more it costs to make necklaces than bracelets. To do this, we will subtract how much it costs to make bracelets from how much it takes to make necklaces. Necklace−(8b+6)−Bracelet(4+5b) We will now simplify this polynomial. (8b+6)−(4+5b)Distribute -18b+6−4−5bCommutative Property of Addition(8b−5b)+(6−4)Add and subtract terms3b+2 The polynomial is 3b+2. | |

Exercises 42 We want to write an expression that represents the total memberships. To do this, we will add the individual memberships with the family memberships. (142+12m)+(52+6m) We will now simplify this polynomial by adding like terms. (142+12m)+(52+6m) Simplify Remove parentheses142+12m+52+6mCommutative Property of Addition(12m+6m)+(142+52)Add terms 18m+194 The polynomial simplifies to 18m+194. | |

Exercises 43 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (2s2−5st−t2)−(s2+7st−t2) In this case, we have two s2-terms, two st-terms, and two t2-terms. All terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (2s2−5st−t2)−(s2+7st−t2)Distribute -1(2s2−5st−t2)−s2−7st+t2Remove parentheses2s2−5st−t2−s2−7st+t2Commutative Property of Addition2s2−s2−5st−7st−t2+t2Subtract terms2−12st−t2+t2Add termss2−12st | |

Exercises 44 The first step in simplifying this expression is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (a2−3ab+2b2)+(-4a2+2ab−b2) In this case, we have two a2-terms, two ab-terms, and two b2-terms. All terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (a2−3ab+2b2)+(-4a2+2ab−b2)Remove parenthesesa2−3ab+2b2−4a2+2ab−b2Commutative Property of Additiona2−4a2−3ab+2ab+2b2−b2Subtract term-3a2−3ab+2ab+b2Add terms-3a2−ab+b2 | |

Exercises 45 The first step in simplifying this expression is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable and the same exponent — can be combined. (c2−6d2)+(c2−2cd+2d2) In this case, we have two c2-terms, one cd-term, and two d2-terms. Only c2-terms and d2-terms can be combined, so to simplify the expression we will rearrange it according to the Commutative Property of Addition and then combine like terms. (c2−6d2)+(c2−2cd+2d2)Remove parenthesesc2−6d2+c2−2cd+2d2Commutative Property of Additionc2+c2−2cd−6d2+2d2Add terms2c2−2cd−4d2 | |

Exercises 46 The first step in finding a difference is to identify which, if any, terms can be combined. Remember, only like terms — constant terms or terms with the same variable — can be combined. (-x2+9xy)−(x2+6xy−8y2) In this case, we have two x2-terms, two xy-terms, and one y2-terms. All terms can be combined. To simplify the expression we will start by using Distributive Property and then we will rearrange it according to the Commutative Property of Addition. (-x2+9xy)−(x2+6xy−8y2)Distribute -1(-x2+9xy)−x2−6xy+8y2Remove parentheses-x2+9xy−x2−6xy+8y2Commutative Property of Addition-x2−x2+9xy−6xy+8y2Subtract term-2x2+3xy+8y2 | |

Exercises 47 Let's start by reviewing the definition of a polynomial. A polynomial is a monomial or a sum or difference of monomials. Since in algebraic expressions we used the + or − signs to separate terms, we can conclude that every term of a polynomial is always a monomial. With this in mind, we can complete the exercise's sentence.Every term of a polynomial is always a monomial. | |

Exercises 48 Let's start by reviewing the definition of a trinomial. A trinomial is a polynomial with exactly 3 terms, each of them being a monomial. Since we can only combine like terms, the difference of two trinomials can have different outcomes. Some examples are shown below.Difference of two trinomialsCombine like termsSimplifyIs it a trinomial? -4x2+8x+7−(3x2+3x+4)(-4−3)x2+(8−3)x+(7−4)-7x2+5x+3Yes 3x3+2x+1−(3x2−2x+4)3x3−3x2+(2+2)x+(1−4)3x3−3x2+4x−3No 3x2−2x+7−(3x2+2x+3)(3−3)x2+(-2−2)x+(7−3)-4x+4No -3x2−2x−7−(-3x2−2x−7)(-3+3)x2+(-2+2)x+(-7+7)0No As we can see, the difference of two trinomials is not necessarily a trinomial as well. This happens just sometimes.The difference of two trinomials is sometimes a trinomial. | |

Exercises 49 Let's start by reviewing the definition of a binomial. A binomial is a polynomial with exactly two terms, each of them being a monomial. Therefore, just as with any other polynomial, the degree of a binomial is the greatest degree of its terms. Some examples are shown below.Example binomialHighest degree monomialDegree of the binomial 5x2+4x5x22 2x5+3x2x55 5x+45x11 With this in mind we can complete the exercise's sentence.A binomial is sometimes a polynomial of degree 2. | |

Exercises 50 Let's start by reviewing what polynomiasl are and how to add them. A polynomial is a monomial or a sum of monomials, each of them a term of the polynomial. We can add polynomials by combining like terms. Therefore, the sum of two polynomials will always be a polynomial as well. | |

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Exercises 52 | |

Exercises 53 | |

Exercises 54 | |

Exercises 55 | |

Exercises 56 | |

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Exercises 58 | |

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Exercises 64 |

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

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Multiplying 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