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One way to analyze a quadratic expression is to factor it — writing it as the product of two binomials. In this lesson, quadratic expressions with a leading coefficient of $1$ will be factored.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

A quadratic trinomial in the form $x_{2}+bx+c$ can be factored as $(x+p)(x+q)$ if there exist $p$ and $q$ such that $p+q=b$ and $pq=c.$

$x_{2}+bx+c=(x+p)(x+q) $

Suppose that the sum of two numbers $p$ and $q$ is equal to $b$ and their product is equal to $c.$
Therefore, the quadratic trinomial and the product of the binomials are equal.

$p+q=bandp⋅q=c $

To show $x_{2}+bx+c=(x+p)(x+q),$ substitute the equivalent expressions for $b$ and $c.$
$x_{2}+bx+c$

SubstituteExpressions

Substitute expressions

$x_{2}+(p+q)x+pq$

Factor

Distr

Distribute $x$

$x_{2}+px+qx+pq$

FactorOut

Factor out $x$

$x(x+p)+qx+pq$

FactorOut

Factor out $q$

$x(x+p)+q(x+p)$

FactorOut

Factor out $x+p$

$(x+q)(x+p)$

CommutativePropMult

Commutative Property of Multiplication

$(x+p)(x+q)$

$x_{2}+bx+c=(x+p)(x+q) $

As an example, the trinomial below will be factored.
*expand_more*
*expand_more*

*expand_more*

$x_{2}+7x+12 $

These three steps can be followed to factor it.
1

Analyze the Signs of $b$ and $c$

For some $p$ and $q,$ the aim is to write the given trinomial as follows.

$x_{2}+7x+12=(x+p)(x+q) $

To do so, the signs of $b$ and $c$ will be used to determine the signs of $p$ and $q.$
$x_{2}+7x+12 $

Here, $b=7$ and $c=12,$ so both $b$ and $c$ are positive. - Since $c$ is positive, the factors $p$ and $q$ must have the same sign so that $p⋅q$ is positive.
- Since $b$ is positive, both $p$ and $q$ must be positive so that $p+q$ is positive.

2

Find the Pair of Factors of $c$ That Has a Sum of $b$

It is known that $b=7,$ $c=12$ and that $p$ and $q$ are positive integers. Therefore, two positive factors of $12$ whose sum is $7$ need to be found. The positive factor pairs of $12$ will be listed and the pair with a sum of $7$ identified.

Positive Factors of $12$ | Sum |
---|---|

$1$ and $12$ | $1+12=13$ |

$2$ and $6$ | $2+6=8$ |

$3$ and $4$ | $3+4=7$ |

As seen, the factor pair of $3$ and $4$ meet these requirements, so the values of $p$ and $q$ are $3$ and $4.$

3

Write in Factored Form

For the given trinomial, two integers with a sum of $7$ and a product of $12$ were found.

$x_{2}+7x+12⇒p=3q=4 $

Therefore, the trinomial can be written as the product of the binomials $x+3$ and $x+4.$
$x_{2}+7x+12=(x+3)(x+4) $

It's a three-day weekend and Tearrik has the day off from school. He wants to use this time to make a present for his brother's birthday. He bought a nice frame and then chose a photo of the two of them. However, the photo does not fit in the frame, so Tearrik needs to edit it.

The editing program represents the area of the photo as $x_{2}+13x+42$ and its length as $x+7.$ Help Tearrik answer the following questions.

a Write a binomial that represents the width of the photo.

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b Tearrik is considering making a new picture frame himself. He needs to know approximately how much material he would need to make the frame. Find the perimeter of the photo if it is $9$ inches wide.

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a Start by factoring the given quadratic expression.

b Use the given information to find $x.$ Then, use the formula for the perimeter of a rectangle.

a The given quadratic expression represents the area of the photo. Therefore, it should be equal to the product of its length $x+7$ and width.

$Ax_{2}+13x+42 =ℓ⋅w⇓=(x+7)w $

To find $w,$ the trinomial must be factored. To do so, identify $b$ and $c,$ and determine the signs of $p$ and $q.$
$x_{2}+13x+42 $

Here, $b=13$ and $c=42,$ so both $b$ and $c$ are positive. - Since $c$ is positive, the factors $p$ and $q$ must have the same sign so that $p⋅q$ is positive.
- Since $b$ is positive, both $p$ and $q$ must be positive so that $p+q$ is positive.

Two **positive** factors of $42$ whose sum is $13$ need to be found. Now, list the positive factor pairs of $42$ and identify the pair with a sum of $13.$

Positive Factors of $42$ | Sum |
---|---|

$1$ and $42$ | $1+42=43$ |

$2$ and $21$ | $2+21=23$ |

$3$ and $14$ | $3+14=17$ |

$6$ and $7$ | $6+7=13$ |

$x_{2}+13x+42=(x+7)(x+6) $

Since $x+7$ represents the length of the photo, the $x+6$ must represent the width of the photo.
b The width of the photo is given as $9$ inches. Since the binomial $x+6$ represents the width of the photo, the value of $x$ can be found.

$x+6=9⇒x=3 $

Before finding the perimeter of the photo, its length $x+7$ must also be computed.
Now, the length and width of the photo are known. $w=9andℓ=10 $

The perimeter can now be calculated by using the perimeter formula.
The photo's perimeter is $38$ inches.
Vincenzo is using the extra day off school to begin building a kennel for his dog. In order to use the garden area in the most effective way, he has to build it on a triangular base. He draws a plan of a right triangle whose hypotenuse is represented by the binomial $x+7.$

It is known that the trinomial $x_{2}+3x−10$ is twice the area of the triangle and that the length of $BC$ is greater than the length of $AB.$

a What are the possible leg lengths of the triangle when $x=10?$

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b Write an expression for the perimeter of the triangle to find the amount of fencing Vincenzo will need.

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a The given trinomial is equal to the product of the leg lengths of the triangle. Factor the given quadratic trinomial.

b The perimeter of a triangle is the sum of all of its side lengths.

a Since the given trinomial is twice the area of the triangle, the trinomial is equal to the product of leg lengths.

$x_{2}+3x−10=2 ⋅2 AB⋅BC $

To find $AB$ and $BC,$ the quadratic trinomial needs to be written as a product of two binomials.
$x_{2}+3x−10=(x+p)(x+q) $

Now identify $b$ and $c$ to get an idea about the signs of $p$ and $q.$
$x_{2}+3x−10⇓x_{2}+3x+(-10) $

For this expression, $b=3$ and $c=-10,$ so $b$ is positive but $c$ is negative. - Since $c$ is negative, the factors $p$ and $q$ must have opposite signs so that $p⋅q$ is negative.
- Since $b$ is positive, the factor with a greater absolute value must be positive.

As such, the factor pairs of $-10$ where one factor is negative should be listed. Then, look for the pair with a sum of $3.$

Positive Factors of $-10$ | Sum |
---|---|

$-1$ and $10$ | $-1+10=9$ |

$1$ and $-10$ | $1+(-10)=-9$ |

$-2$ and $5$ | $-2+5=3$ |

$2$ and $-5$ | $2+(-5)=-3$ |

$x_{2}+3x−10=(x−2)(x+5) $

These two binomials represent the lengths of the legs. When $x=10,$ the expressions will be $8$ and $15.$ $x−2x+5 x=10 8x=10 15 $

This means that the longer side is $15$ and the shorter side is $8.$ Since $BC$ is greater than $AB,$ $BC=15$ and $AB=8.$
b The binomials representing the lengths of the legs are $x−2$ and $x+5.$ Additionally, $x+7$ represents the hypotenuse. To find the perimeter in terms of $x,$ all these binomials should be added.

$P =(x−2)+(x+5)+(x+7)=3x+10 $

Tadeo is using the extra day off school to catch up on homework. He is almost finished with his math homework but is stuck on one problem. He reviews the notes he wrote about factoring the given quadratic trinomial with a leading coefficient of $1.$

Describe the error in his notes and help him factor the trinomial correctly so he can spend the rest of the weekend doing something more fun.**Error:** When $∣p∣>∣q∣,$ $q$ must be a positive integer and $p$ must be a negative integer so that their sum is negative.

**Factored Form:** $(x−15)(x+6)$

Start by identifying the values of $b$ and $c$ for the given quadratic trinomial.

First, $b$ and $c$ will be identified.

The factors that satisfy the conditions are $-15$ and $6,$ so the trinomial can be written as the product of the binomials $x−15$ and $x+6.$

$x_{2}−9x−90⇓x_{2}+(-9)x+(-90) $

Here, $b=-9$ and $c=-90,$ meaning that both $b$ and $c$ are negative. - Since $c$ is negative, the factors $p$ and $q$ must have
**opposite signs**so that $p⋅q$ is negative. - Since $b$ is negative, the factor with a greater absolute value must be negative so that their sum is negative.

Therefore, Tadeo's first point is correct. If it is assumed that $∣p∣>∣q∣,$ then $p$ should be negative. In other words, Tadeo's second point is incorrect.

The given trinomial can be factored using this information. To do so, the factor pairs of $-90$ where one factor is negative and its absolute value is greater than the other factor are listed. Then, the pair with a sum of $-9$ should be looked for.

Factors of $-90$ | Sum of Factors |
---|---|

$-90$ and $1$ | $-90+1=-89$ |

$-45$ and $2$ | $-45+2=-43$ |

$-30$ and $3$ | $-30+3=-27$ |

$-18$ and $5$ | $-18+5=-13$ |

$-15$ and $6$ | $-15+6=-9$ |

$x_{2}−9x−90=(x−15)(x+6) $

An inter-class quiz game is being held at Davontay's school this weekend, and he is his class's champion. The quizmaster Paulina asks Davontay to write a quadratic trinomial and then factor it. The conversation between the quizmaster and Dovantoy is shown in the diagram.

Help Davontay find the value of $b$ and write the trinomial in factored form to win the quiz game.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69444em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">b<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["-16"]}}

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Davontay needs to write a quadratic trinomial in the form $x_{2}+bx+c.$ It is also given that $b$ is negative and $c=60.$

Of these possible values, $-16$ is the greatest, which meets Paulina's hint. Finally, the trinomial can be written.

$x_{2}+bx+c⇓x_{2}+bx+60 $

Since the above trinomial can be factored, the value of $b$ should be the sum of the two factors, $p$ and $q$, of $60.$
$x_{2}+bx+60=(x+p)(x+q) $

Now, two facts can be inferred from this trinomial. - Since $c$ is positive, the factors $p$ and $q$ must have the same sign so that $p⋅q$ is positive.
- Since $b$ is negative, both $p$ and $q$ should be negative so that $p+q$ is negative.

Based on this information, only negative factor pairs of $60$ need to be listed.

Negative Factors of $60$ |
---|

$-1$ and $-60$ |

$-2$ and $-30$ |

$-3$ and $-20$ |

$-4$ and $-15$ |

$-5$ and $-12$ |

$-6$ and $-10$ |

The sum of the factors in each pair could be the value of $b,$ so Davontay's concern is valid. There are six values for $b.$ These values can be found as follows.

Factors of $60$ | Sum |
---|---|

$-1$ and $-60$ | $-1+(-60)=-61$ |

$-2$ and $-30$ | $-2+(-30)=-32$ |

$-3$ and $-20$ | $-3+(-20)=-23$ |

$-4$ and $-15$ | $-4+(-15)=-19$ |

$-5$ and $-12$ | $-5+(-12)=-17$ |

$-6$ and $-10$ | $-6+(-10)=-16$ |

$x_{2}−16x+60 $

The factors $-6$ and $-10$ have a product of $60$ and a sum of $-16.$ Therefore, the trinomial can be written as the product of the binomials $x−6$ and $x−10.$
$x_{2}−16x+60=(x−6)(x−10) $

Factor each quadratic expression with a leading coefficient $1.$ Write the answer in such a way that the value of $q$ is the greater factor.

Maya and her father spent the long weekend building a trough for the animals on their farm. Her father knows that Maya has a math test coming up soon, so he decides to help her prepare for it by quizzing her about the trough they just built.

The trough's length is $105$ centimeters longer than its width. The area covered by the trough is $12250$ square centimeters.

b State if the solutions make sense. What are the width and length of the trough?

a **Equation:** $w(w+105)=12250$

**Solutions:** $w=70$ and $w=-175$

b Only the positive solution makes sense because measures cannot be negative.
**Length:** $175$ centimeters

**Width:** $70$ centimeters

a Use the formula for the area of a rectangle to write an equation.

b Length cannot be negative.

a The width of the trough is represented by $w.$ Since the length of the rectangular trough is $105$ centimeters longer than its width, $w+105$ represents the length.

External credits: Colin Smith

$A=w⋅ℓ⇓12250=w(w+105) $

Now, the equation needs to be rewritten so that a quadratic trinomial is formed on one side of the equation. Then, it will be solved by factoring.
$12250=w(w+105)$

Rewrite

Distr

Distribute $w$

$12250=w_{2}+105w$

SubEqn

$LHS−12250=RHS−12250$

$0=w_{2}+105w−12250$

RearrangeEqn

Rearrange equation

$w_{2}+105w−12250=0$

$w_{2}+105w+(-12250) $

For this trinomial, $b=105$ and $c=-12250.$ Since the value of $c$ is negative, only factor pairs of $-12250$ that have opposite signs will be considered. Since $b$ is positive, the factor with a greater absolute value must be positive. Factors of $-12250$ | Sum of Factors |
---|---|

$1225$ and $-10$ | $1225+(-10)=1215$ |

$490$ and $-25$ | $490+(-25)=465$ |

$350$ and $-35$ | $350+(-35)=315$ |

$245$ and $-50$ | $245+(-50)=195$ |

$175$ and $-70$ | $175+(-70)=105$ |

$w_{2}+105w−12250=0⇓(w−70)(w+175)=0 $

The left-hand side of the equation is a product of two factors. One of them must be zero so that the product is equal to zero. This is known as the Zero Product Property. $(w−70)(w+175)=0$

ZeroProdProp

Use the Zero Product Property

$w−70=0w+175=0 (I)(II) $

AddEqn

$(I):$ $LHS+70=RHS+70$

$w=70w+175=0 $

SubEqn

$(II):$ $LHS−175=RHS−175$

$w=70w=-175 $

b Recall the solutions of the equation written in Part A.

$w=70andw=-175 $

Since $w$ represents a length, it cannot be negative. This means that only the positive value makes sense. The dimensions of the trough can be found by substituting $w=75.$ Expression | $w=75$ | |
---|---|---|

Width | $w$ | $75$ |

Length | $w+105$ | $75+105=175$ |

The width of the trough is $70$ centimeters and the length is $175$ centimeters.

The challenge presented at the beginning can now be solved using the information covered in this lesson. Jordan wants to find a pair of integers with a sum of $11$ and a product of $28.$
### Answer

### Solution

$p+q=11p⋅q=28 $

Help Jordan find the pair of numbers.
$4$ and $7$

The first step is to isolate $q$ in the first equation.
To solve this equation, the quadratic trinomial should be factored.

The factors $-7$ and $-4$ has a sum of $-11$ and a product of $28.$ The trinomial can now be written as the product of the binomials $p−7$ and $p−4.$

$p+q=11⇒q=11−p $

Now, the expression equivalent to $q$ is substituted into the other equation. Then, the equation is rewritten so that a quadratic expression is formed on one side of the equation.
$p⋅q=28$

Substitute

$q=11−p$

$p⋅(11−p)=28$

Rewrite

Distr

Distribute $p$

$11p−p_{2}=28$

SubEqn

$LHS−28=RHS−28$

$11p−p_{2}−28=0$

MultEqn

$LHS⋅(-1)=RHS⋅(-1)$

$-11p+p_{2}+28=0$

CommutativePropAdd

Commutative Property of Addition

$p_{2}−11p+28=0$

$p_{2}−11p+28 $

In this trinomial, $b=-11$ and $c=28.$ Since $c$ is positive, factor pairs of $28$ that have the same sign should be considered. Of those pairs, negative pairs are listed because $b$ is negative. Factors of $28$ | Sum of Factors |
---|---|

$-28$ and $-1$ | $-28+(-1)=-29$ |

$-14$ and $-2$ | $-14+(-2)=-16$ |

$-7$ and $-4$ | $-7+(-4)=-11$ |

$p_{2}−11p+28=0⇓(p−7)(p−4)=0 $

The left-hand side of the equation is a product of two factors. One of them must be zero so that the product is equal to zero. Therefore, $p$ is either $7$ or $4$ by the Zero Product Property.
$(p$