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Some of the easiest quadratic expressions to factor are those with a leading coefficient of $1.$ This lesson will explain how to factor quadratic expressions with leading coefficients other than $1.$
### Catch-Up and Review

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

Consider the numbers $a=2,$ $b=12,$ and $c=10.$ Find two numbers $p$ and $q$ such that the following conditions are satisfied.

$Condition1pq=ab Condition2p+q=ac $

If the greatest common factor (GCF) of a quadratic trinomial of the form $ax_{2}+bx+c$ is $a,$ then it can be rewritten by factoring out $a.$ In this case, since $a$ is a factor of both $b$ and $c,$ these two coefficients can be written in terms of $a.$
The leading coefficient of the quadratic expression in the parentheses is $1.$ Therefore, it can be factored by finding two numbers $p$ and $q$ such that $p+q=b_{1}$ and $pq=c_{1}.$

$bc =a⋅b_{1},for some numberb_{1}=a⋅c_{1},for some numberc_{1} $

With this information in mind, the expression $ax_{2}+bx+c$ can be factored.
$ax_{2}+bx+c$

SubstituteII

$b=a⋅b_{1}$, $c=a⋅c_{1}$

$ax_{2}+a⋅b_{1}x+a⋅c_{1}$

FactorOut

Factor out $a$

$a(x_{2}+b_{1}x+c_{1})$

$a(x_{2}+b_{1}x+c_{1})=a(x+p)(x+q)⇕ax_{2}+bx+c=a(x+p)(x+q) $

Magdalena's class is planning a camping trip. Magdalena volunteered to research tents to purchase. When she asked her teacher what size they would buy, he instead told her that the base area of the tent should be $3x_{2}+12x+9$ square feet and its width $w$ is $x+3$ feet.

To choose a suitable tent, Magdalena needs to know the dimensions of the tent's base.

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b What will be the dimensions of the tent when $x=1$ foot?

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a Think about the greatest common factor of the given quadratic trinomial.

b Substitute the given value into the expressions that represent the width and length of the tent.

$A=wℓ⇓3x_{2}+12x+9=(x+3)ℓ $

To find an expression for the length $ℓ,$ the quadratic trinomial that represents the area will be factored. The first step to factoring the polynomial is to find its greatest common factor (GCF), if there is one. Note that the GCF of this expression is $3.$
$3x_{2}+12x+9$

SplitIntoFactors

Split into factors

$3x_{2}+3(4)x+3(3)$

FactorOut

Factor out $3$

$3(x_{2}+4x+3)$

Factors of $3$ | Sum of the Factors |
---|---|

$-1$ and $-3$ | $-1+(-3)=-4×$ |

$1$ and $3$ | $1+3=4✓$ |

$3(x_{2}+4x+3)⇕3(x+1)(x+3) $

This expression represents the area of a tent's base.
$A=wℓ⇓3(x+1)(x+3)=(x+3)ℓ $

The expression $(x+3)$ on the right-hand side of the equation represents the width of the tent's base. A measurement for a width cannot be zero, which means that both sides of the equation can be divided by $(x+3).$
$3(x+1)(x+3)=(x+3)ℓ$

DivEqn

$LHS/(x+3)=RHS/(x+3)$

$x+33(x+1)(x+3) =x+3(x+3)ℓ $

CancelCommonFac

Cancel out common factors

$x+3 3(x+1)(x+3) =x+3 (x+3) ℓ $

SimpQuot

Simplify quotient

$3(x+1)=ℓ$

RearrangeEqn

Rearrange equation

$ℓ=3(x+1)$

Distr

Distribute $3$

$ℓ=3x+3$

b Now that the equations for the width and length of the tent, substitute $x=1$ into each binomial to find the corresponding values.

Substitution | Calculation | |
---|---|---|

$w=x+3$ | $w=1+3$ | $w=4$ feet |

$ℓ=3x+3$ | $ℓ=3⋅1+3$ | $ℓ=6$ feet |

The class will buy tents that are $4$ feet wide and $6$ feet long.

When trying to factor a quadratic trinomial of the form $ax_{2}+bx+c,$ it can be difficult to *expand_more*
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seeits factors. Consider the following expression.

$8x_{2}+26x+6 $

Here, $a=8,$ $b=26,$ and $c=6.$ There are six steps to factor this trinomial.
1

Factor Out the GCF of $a,$ $b,$ and $c$

To fully factor a quadratic trinomial, the Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ has to be factored out first. To identify the GCF of these numbers, their prime factors will be listed.
In the remaining steps, the factored coefficient $2$ before the parentheses can be ignored. The new considered quadratic trinomial is $4x_{2}+13x+3.$ Therefore, the current values of $a,$ $b,$ and $c,$ are $4,$ $13,$ and $3,$ respectively. If the GCF of the coefficients is $1,$ this step can be ignored.

$8=26=6= 2⋅2⋅22⋅132⋅3 $

It can be seen that $8,$ $26,$ and $6$ share exactly one factor, $2.$
$GCF(8,26,6)=2 $

Now, $2$ can be factored out.
$8x_{2}+26x+6$

SplitIntoFactors

Split into factors

$2(4)x_{2}+2(13)x+2(3)$

FactorOut

Factor out $2$

$2(4x_{2}+13x+3)$

2

Find the Factor Pair of $ac$ Whose Sum Is $b$

It is known that $a=4$ and $c=3,$ so $ac=12>0.$ Therefore, the factors must have the same sign. Also, $b=13.$ Since the sum of the factors is positive and they must have the same sign, both factors must be positive. All positive factor pairs of $12$ can now be listed and their sums checked.

Factors of $ac$ | Sum of Factors |
---|---|

$1$ and $12$ | $1+12=13✓$ |

$2$ and $6$ | $2+6=8×$ |

$3$ and $4$ | $3+4=7×$ |

In this case, the correct factor pair is $1$ and $12.$ The following table sums up how to determine the signs of the factors based on the values of $ac$ and $b.$

$ac$ | $b$ | Factors |
---|---|---|

Positive | Positive | Both positive |

Positive | Negative | Both negative |

Negative | Positive | One positive and one negative. The absolute value of the positive factor is greater. |

Negative | Negative | One positive and one negative. The absolute value of the negative factor is greater. |

Such analysis makes the list of possible factor pairs shorter.

3

Write $bx$ as a Sum

The factor pair obtained in the previous step will be used to rewrite the $x-$term — the linear term — of the quadratic trinomial as a sum. Remember that the factors are $1$ and $12.$
The linear term $13x$ can be rewritten in the original expression as $12x+x.$

$4x_{2}+13x+3⇕4x_{2}+12x+x+3 $

4

Factor Out the GCF of the First Two Terms

The expression has four terms, which can be grouped into the $first$ $two$ and the $last$ $two$ $terms.$ Then, the GCF of each group can be factored out.

$4x_{2}+12x+x+3 $

The first two terms, $4x_{2}$ and $12x,$ can be factored.
$4x_{2}=12x= 2⋅2⋅x⋅x2⋅2⋅3⋅x $

The GCF of $4x_{2}$ and $12x$ is $2⋅2⋅x=4x.$
5

Factor Out the GCF of the Last Two Terms

6

Factor Out the Common Factor

If all the previous steps have been performed correctly, there should now be two terms with a common factor.

$4x(x+3)+1(x+3) $

The common factor will be factored out.
The factored form of $4x_{2}+13x+3$ is $(4x+1)(x+3).$ Remember that the original trinomial was $8x_{2}+26x+6$ and that the GCF $2$ was factored out in Step $1.$ This GCF has to be included in the final result.
$8x_{2}+26x+6=2(4x+1)(x+3) $

Heichi and Maya are reading a pamphlet about local campsites. They volunteered to find a suitable campground for their class field trip. Their math teacher, who goes camping often, gives them a challenge to help them decide where to go. If they answer his questions about the pamphlet correctly, he will tell them about his favorite local campsite.

Their teacher tells them that the area of the pamphlet is $3x_{2}−5x+2$ square inches and that it has a width of $x−1$ inches.

a Write a binomial that represents the length of the pamphlet.

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b Find the perimeter of the pamphlet if the width is $4$ inches.

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$A=wℓ⇓3x_{2}−5x+2=(x−1)ℓ $

To find an expression for the length $ℓ,$ the trinomial that represents the area will be factored. First, the values of $a,$ $b,$ and $c$ for the trinomial will be determined.
$3x_{2}−5x+2⇕3x_{2}+(-5)x+2 $

Here, $a=3,$ $b=-5,$ and $c=2.$ With this in mind, the following conclusions can be made. - The Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ is $1.$
- Since $ac=6,$ which is greater than zero, the factors must have the same sign.
- Since $b=-5,$ which is negative, the sum of the factors will be negative. Therefore, both of the factors must be negative.

Now, all the negative factor pairs of $6$ will be listed and checked whether their sum is $-5.$

Factors of $ac=6$ | Sum of Factors |
---|---|

$-1$ and $-6$ | $-1+(-6)=-7$ |

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

$3x_{2}−5x+2⇕3x_{2}−2x−3x+2 $

Next, the expression will be factored by grouping its terms.
$3x_{2}−2x−3x+2$

CommutativePropAdd

Commutative Property of Addition

$3x_{2}−3x−2x+2$

$3x⋅x−3x⋅1−2x+2$

FactorOut

Factor out $3x$

$3x(x−1)−2x+2$

Rewrite

Rewrite $2$ as $2⋅1$

$3x(x−1)−2x+2⋅1$

FactorOut

Factor out $-2$

$3x(x−1)−2(x−1)$

FactorOut

Factor out $(x−1)$

$(3x−2)(x−1)$

$3x_{2}−5x+2=(x−1)(3x−2) $

Since $x−1$ represents the width of the pamphlet, the binomial $3x−2$ represents its length.
b To find the perimeter of a rectangle, all four side lengths must be added. The width of the pamphlet is given as $4$ inches. Since the binomial $x−1$ represents the width, the value of $x$ can be found from here.

$x−1=4⇔x=5 $

Then, the length of the pamphlet will be found by substituting $x=5$ into the binomial that it was found in Part A.
The length of the pamphlet is $13$ inches. The perimeter of the pamphlet can finally be calculated now that both the length and width are known.
The perimeter of the pamphlet is $34$ inches.
At the campsite, each camper is assigned a rectangular area of $28$ square feet for their tent. Dominika wants to place a tarp on the ground below her tent to make sure it stays dry underneath. She does not remember the size of the tarp she has, but she knows that its length is $1$ inch less than twice its width. *exactly* in the allocated area. Help her find the length and width of the tarp. ### Hint

### Solution

Now, the obtained quadratic expression will be factored. First, its coefficients will be identified.

Of these factors, $-8$ and $7$ satisfy the conditions. Therefore, the $x-$term of the trinomial on the left-hand side of the equation will be rewritten by using these numbers as a sum.
Since the product of the two binomials is zero, at least one binomial must equal zero. With this in mind, the Zero Product Property will be applied.
The solutions to the equation are $4$ and $-3.5.$ Since a negative length does not make sense, $w$ cannot be negative. This means that the value of the width is $4$ feet. With this information and knowing that the length is $1$ foot less than twice the width, the length can be found.
The base of the tarp has a length and a width of $7$ and $4$ feet, respectively.

Dominika finds that the tarp fits

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Start by writing a quadratic expression for the area of the tarp. Then, factor the obtained expression.

An expression for the area of the tarp can be written by using the expressions for its length and width. Let $w$ be the width of the tarp. Since the length is $1$ inch less than twice the width, it can be expressed as $2w−1.$

Since the base is a rectangle, its area is equal to the product of its width and length.$A=w ℓ$

SubstituteII

$A=28$, $ℓ=2w−1$

$28=w(2w−1)$

Rewrite

Distr

Distribute $w$

$28=2w_{2}−w$

SubEqn

$LHS−28=RHS−28$

$0=2w_{2}−w−28$

RearrangeEqn

Rearrange equation

$2w_{2}−w−28=0$

$2w_{2}−w−28=0⇕2w_{2}+(-1)w+(-28)=0 $

Here, $a=2,$ $b=-1,$ and $c=-28.$ With this in mind, the following conclusions can be made. - The Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ is $1.$
- Since $ac=-56,$ which is less than zero, the factors must have the different signs. Therefore, there is a positive factor and a negative factor.
- Since $b=-1,$ the sum of the factors is negative. This means that the absolute value of the negative factor is greater than the absolute value of the positive factor.

Now, the factor pairs of $-56$ with opposite signs and where the absolute value of the negative factor is greater than the absolute value of the positive number will be listed. Also, their sum will be calculated to see if it is $-1.$

Factors of $ac=-56$ | Sum of Factors |
---|---|

$-56$ and $1$ | $-56+1=-55$ |

$-28$ and $2$ | $-28+2=-26$ |

$-14$ and $4$ | $-14+4=-10$ |

$-8$ and $7$ | $-8+7=-1$ |

$2w_{2}−w−28=0⇓2w_{2}−8w+7w−28=0 $

Then, the obtained expression will be factored by grouping.
$2w_{2}−8w+7w−28=0$

$2w⋅w−2w⋅4+7w−28=0$

FactorOut

Factor out $2w$

$2w(w−4)+7w−28=0$

SplitIntoFactors

Split into factors

$2w(w−4)+7w−7⋅4=0$

FactorOut

Factor out $7$

$2w(w−4)+7(w−4)=0$

FactorOut

Factor out $(w−4)$

$(2w+7)(w−4)=0$

$(w−4)(2w+7)=0$

Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

$w−4=02w+7=0 (I)(II) $

AddEqn

$(I):$ $LHS+4=RHS+4$

$w=42w+7=0 $

SubEqn

$(II):$ $LHS−7=RHS−7$

$w=42w=-7 $

DivEqn

$(II):$ $LHS/2=RHS/2$

$w=4w=-3.5 $

$Width:Length: 4feet7feet $

Dylan is practicing on the basketball court at the campground. For the ball to fall through the hoop without hitting the rim or backboard, it needs to follow a specific path after being thrown at a $45_{∘}$ angle.
### Hint

### Solution

This path can be modeled with a quadratic function.

$h(t)=-5t_{2}+13t+6 $

Here, $h$ is the height of the ball in feet $t$ seconds after it has been thrown. Using this model, find the time it takes for the ball to hit the ground after Dylan throws it. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Time:","formTextAfter":"seconds","answer":{"text":["3"]}}

The value of $h$ is zero when the ball hits the ground.

The height of the ball is zero when the it hits the ground. Therefore, to find the total time it takes for the ball to hit the ground, $0$ will be substituted for $h(t)$ in the given equation.

The factors of $-30$ whose sum is $13$ are $-2$ and $15.$ The $t-$term of the right-hand side can be rewritten as a subtraction of these numbers.

$h(t)=-5t_{2}+13t+6⇓0=-5t_{2}+13t+6 $

Now, the trinomial will be factored. First, its coefficients will be identified.
$-5t_{2}+13t+6 $

Here, $a=-5,$ $b=13,$ and $c=6.$ With this information in mind, the following conclusions can be made. - The Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ is $1.$
- Since $ac=-30,$ which is less than zero, the factors must have different signs. Therefore, one of the factors is positive and one of the factors is negative.
- Since $b=13,$ the sum of the factors is positive. Therefore, the absolute value of the positive factor is greater than the absolute value of the negative factor.

Now, the factor pairs of $-30$ with opposite signs and where the absolute value of the positive factor is greater than the absolute value of the negative factor will be listed. Also, their sum will be calculated to see whether it is $13.$

Factors of $ac=-30$ | Sum of Factors |
---|---|

$-1$ and $30$ | $29$ |

$-2$ and $15$ | $13$ |

$-3$ and $10$ | $7$ |

$-5$ and $6$ | $1$ |

$0=-5t_{2}+13t+6⇓0=-5t_{2}+15t−2t+6 $

Now, the obtained expression will be factored by grouping.
$0=-5t_{2}+15t−2t+6$

RearrangeEqn

Rearrange equation

$-5t_{2}+15t−2t+6=0$

$-5t⋅t+5t⋅3−2t+6=0$

FactorOut

Factor out $-5t$

$-5t(t−3)−2t+6=0$

Rewrite

Rewrite $6$ as $2⋅3$

$-5t(t−3)−2t+2⋅3=0$

FactorOut

Factor out $-2$

$-5t(t−3)+(-2)(t−3)=0$