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Quadratic equations can be solved using square roots. Sometimes this method might be inconvenient to use for equations of the form $ax_{2}+bx+c=0.$ The following lesson introduces another method of solving such equations.
### Catch-Up and Review

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

Challenge

Consider the following quadratic equation for $x.$

$2x_{2}+2x−4=0 $

Can this equation be solved by taking square roots? If not, how can it be solved algebraically — without graphing?Discussion

An expression written in factored form and set equal to $0$ can be solved using the Zero Product Property. When the product of two or more factors is $0,$ at least one of the factors must equal $0.$ Consider the following equation.
*expand_more*
*expand_more*

$(3x−9)(x+5)=0 $

Since the left-hand side of the equation is in factored form, there are two steps to solve it.
1

Set Each Factor Equal to $0$

By the Zero Product Property, at least one of the factors must equal $0.$ Therefore, each of the factors can be set equal to $0.$
**or**. By setting the factors equal to $0,$ new equations are created. There are as many equations as there were factors — in this case there are two.

$3x−9=0orx+5=0 $

Note that the word connecting the equations is 2

Solve the Obtained Equations

Now, inverse operations will be used to solve the new equations.
The solutions to the new equations **both** solve the original equation. Therefore, $x=3$ and $x=-5$ are the solutions to $(3x−9)(x+5)=0.$

Pop Quiz

The following applet shows a quadratic equation whose left-hand side is written in factored form. Using the Zero Product Property find the solutions of the equation. Round to the nearest tenth if necessary.

Discussion

A quadratic equation in the form $ax_{2}+bx=0$ is obtained when the constant term $c$ is equal to $0.$ Such quadratic equations can be solved using the Zero Product Property. Consider the following equation.
*expand_more*
*expand_more*
*expand_more*

$3x_{2}−12x=0 $

There are three steps to solve it.
1

Factor Out the GCF

First, the greatest common factor (GCF) of $ax_{2}$ and $bx$ has to be factored out. In the considered case, $ax_{2}=3x_{2}$ and $bx=-12x.$ The factors of these two terms will be now listed.

$3x_{2}=-12x= 3⋅x⋅x-4⋅3⋅x $

The GCF is $3⋅x=3x.$ Finally, it can be factored out.
2

Set Each Factor Equal to $0$

After factoring out the GCF, the left-hand side of the equation is in factored form. Now, the steps of solving an equation using the Zero Product Property will be followed. By the Zero Product Property each of the factors can be set equal to $0.$

$3x(x−4)=0⇒3x=0x−4=0 (I)(II) $

3

Solve the Obtained Equations

The two linear equations obtained in the previous step will be now solved using inverse operations.
The solutions to the original equation are $x=0$ and $x=4.$ Please note that $x=0$ makes the expression $ax_{2}+bx$ equal to $0$ for any values of $a$ and $b.$ Therefore, it is a solution to any equation of the form $ax_{2}+bx=0.$

Example

Dolphins jump out of the water to improve their navigation and to see the surface of the ocean. They also do it for fun.

The following equation models the height $h$ in meters of a jumping dolphin $t$ seconds after it leaves the water.$h=-5t_{2}+8t $

After how many seconds is the dolphin back in the water? Write and solve by factoring the appropriate quadratic equation. {"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":"The dolphin is back in the water after","formTextAfter":"seconds.","answer":{"text":["1.6","\\dfrac{16}{10}","\\dfrac{8}{5}","1\\dfrac{6}{10}","1\\dfrac{3}{5}"]}}

The dolphin is in the water when $h=0.$

When $h=0,$ the dolphin is in the water. An equation for $t$ can be written by substituting $0$ for $h$ in the given equation.
The left-hand side of the equation is written in factored form and the right-hand side is equal to $0.$ Therefore, this equation can be solved using the Zero Product Property.
The solutions are $t=0$ and $t=1.6.$ It is known that at $t=0$ the dolphin is still in the water. Therefore, it must be back in the water after $1.6$ seconds.

$0=-5t_{2}+8t $

The solutions of the above equation represent the time when the dolphin leaves the water and the time when the dolphin is back in the water. The equation will be now solved. First, it will be rearranged, as it is usually preferred to have the variable terms on the left-hand side.
$0=-5t_{2}+8t⇔-5t_{2}+8t=0 $

Next, the greatest common factor (GCF) of the terms on the left-hand side will be factored out. The factors of $-5t_{2}$ and $8t$ are listed below.
$-5t_{2}=8t= -1⋅5⋅t⋅t2⋅2⋅2⋅t $

The two terms have only one common factor $t.$ Therefore, the GCF of $-5t_{2}$ and $8t$ is $t.$
$-5t_{2}+8t=0$

Rewrite

Rewrite $-5t_{2}$ as $t(-5t)$

$t(-5t)+8t=0$

CommutativePropMult

Commutative Property of Multiplication

$t(-5t)+t⋅8=0$

FactorOut

Factor out $t$

$t(-5t+8)=0$

Discussion

Method

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

*expand_more*
*expand_more*
*expand_more*
*expand_more*

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) $

The above method can be used to factor any quadratic trinomial. It is particularly useful for trinomials in the forms $ax_{2}+bx$ or $ax_{2}+c.$ However, in these two cases there are other approaches that might be used as well.

- A quadratic trinomial is of the form $ax_{2}+bx$ when the constant term $c$ is equal to $0.$ To write this expression in factored form, the GCF of $ax_{2}$ and $bx$ has to be factored out.
- A quadratic trinomial is of the form $ax_{2}+c$ when $b$ is equal to $0.$ Sometimes such expressions can be factored using the Difference of Squares formula.

Please note that not every quadratic trinomial can be factored. When a pair of integers whose product is equal to $a⋅c$ and whose sum is $b$ cannot be found, the trinomial *cannot* be factored using the described method.

$x_{2}−x+1,a⋅c=1,b=-1$ | |
---|---|

Product of Factors | Sum of Factors |

$1⋅1=1$ | $1+1=2×$ |

$-1(-1)=1$ | $-1+(-1)=-2×$ |

$ax_{2}+bx+c=0 $

If its left-hand side is factorable, the equation can be solved using the Zero Product Property.Example

Zosia is a student studying how many Dolphins swim past a certain area of cove. She marks off a rectangle using rope where she will count how many dolphins pass through in any given day. The study area's rectangle is equal to $42$ square meters. However, the dimensions of the rectangle are unknown.

What are the width and the length of the rectangle?{"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":"Width:","formTextAfter":null,"answer":{"text":["6"]}}

{"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":"Length:","formTextAfter":null,"answer":{"text":["7"]}}

Based on the given information, write and solve a quadratic equation for $x.$

It is known that the area of the given rectangle is $42m_{2}.$
The equation is written in the form $ax_{2}+bx+c=0.$ The values of $a,$ $b,$ and $c$ can be identified.
*greater than* the absolute value of the positive factor. The sum of the factors should be $-18.$

The pair of factors whose product is equal to $-280$ and whose sum is $-18$ is $10$ and $-28.$ Now, the linear term can be rewritten as a difference.
Next, the GCF of the first two terms and the last two terms will be factored out. For more information on this process, read about finding the GCF.
Note that for the factorization of the polynomial to work, -7 and not the GCF 7 has to be factored out from the last two terms. Finally, the common factor $4x+5$ will be factored out.
Since the left-hand side of the equation is written in factored form and the right-hand side is $0,$ the equation can be solved using the Zero Product Property.
Two solutions to the quadratic equation have been found. They will now be substituted into the expressions for the width and length of the rectangle.

$Area=42m_{2} $

The area of a rectangle can be calculated by multiplying its dimensions — width and length. In this case, the width and the length of the rectangle are given as algebraic expressions. The product of $2x−1$ and $4x−7$ represents the area of the rectangle.
$(2x−1)(4x−7)=42 $

Note that both dimensions of the rectangle are given in meters, so there is no need to change the units. The actual dimensions of the rectangle can be determined by solving the above equation. First, the equation will be rewritten in standard form by using the Distributive Property and Subtraction Property of Equality.
$(2x−1)(4x−7)=42$

Distr

Distribute $(4x−7)$

$2x(4x−7)−1(4x−7)=42$

Distr

Distribute $2x&-1$

$8x_{2}−14x−4x+7=42$

SubTerm

Subtract term

$8x_{2}−18x+7=42$

SubEqn

$LHS−42=RHS−42$

$8x_{2}−18x−35=0$

$8x_{2}−18x−35=0⇕8x_{2}+(-18)x+(-35)=0 $

It can be seen that $a=8,$ $b=-18,$ and $c=-35.$ Next, the quadratic trinomial on the left-hand side will be written in factored form following the general method. Note that $a,$ $b,$ and $c$ have a greatest common factor (GCF) of $1.$ Next, the product of $a$ and $c$ will be found so that all pairs of integers whose product is equal to $a⋅c$ can be listed.
$a⋅c=8(-35)=-280 $

The product $a⋅c=-280$ and $b=-18$ are both negative. Therefore, one of the factors has to be positive and one negative. Additionally, the absolute value of the negative factor has to be Pair of Factors | Sum of Factors |
---|---|

$1$ and $-280$ | $1+(-280)=-279×$ |

$2$ and $-140$ | $2+(-140)=-138×$ |

$4$ and $-70$ | $4+(-70)=-66×$ |

$5$ and $-56$ | $5+(-56)=-51×$ |

$7$ and $-40$ | $7+(-40)=-33×$ |

$8$ and $-35$ | $8+(-35)=-27×$ |

$10$ and $-28$ | $10+(-28)=-18✓$ |

$14$ and $-20$ | $14+(-20)=-6×$ |

$8x_{2}+10x−28x−35=0$

▼

Factor out $2x&-7$

SplitIntoFactors

Split into factors

$2x⋅4x+2x⋅5−7⋅4x−7⋅5=0$

FactorOut

Factor out $2x$

$2x(4x+5)−7⋅4x−7⋅5=0$

FactorOut

Factor out $-7$

$2x(4x+5)−7(4x+5)=0$

$(4x+5)(2x−7)=0$

ZeroProdProp

Use the Zero Product Property

$4x+5=02x−7=0 (I)(II) $

▼

$(I):$ Solve for $x$

SubEqn

$(I):$ $LHS−5=RHS−5$

$4x=-52x−7=0 $

DivEqn

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

$x=4-5 2x−7=0 $

MoveNegNumToFrac

$(I):$ Put minus sign in front of fraction

$x=-45 2x−7=0 $

CalcQuot

$(I):$ Calculate quotient

$x=-1.252x−7=0 $

$x=-1.25x=3.5 $

$2x−1$ | $4x−7$ | |
---|---|---|

$x=-1.25$ | $2(-1.25)−1=-3.5×$ | $4(-1.25)−7=-12×$ |

$x=3.5$ | $2(3.5)−1=6✓$ | $4(3.5)−7=7✓$ |

It can be seen that $x=-1.25$ results in a negative width and length, which is not logical for the dimensions of a rectangle. Therefore, the only valid solution is $x=3.5.$ The width of the rectangle is $6$ meters and the length is $7$ meters.

Example

Magdalena came to help Zosia in her study. Both were trying to figure out the following quadratic equation.
### Hint

### Solution

$9x_{2}−30x+30=5 $

They want to solve the equation by factoring. Magdalena found that the equation has one solution. Zosia determined that it has no solutions. Who is correct? {"type":"choice","form":{"alts":["Magdalena","Zosia"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}

Rewrite the equation in standard form. Factor the left-hand side of the rewritten equation.

To decide which girl is correct, the given equation will be solved by factoring. First, it will be rewritten in standard form.
In this case $a=9,$ $b=-30,$ and $c=25.$ The greatest common factor (GCF) of these numbers is $1.$ Now, all pairs of integer factors whose product is equal to $a⋅c=225$ will be listed. Since $a⋅c$ is positive and $b$ is negative, both factors have to be negative. Their sum will be also calculated in the table, it should be equal to $-30.$

It can be noted that the pair of factors whose product is equal to $a⋅c=225$ and whose sum is $b=-30$ is $-15$ and $-15.$ The linear term on the left-hand side will be rewritten as a difference and then the trinomial will be factored.

Pair of Factors | Sum of Factors |
---|---|

$-1$ and $-225$ | $-1+(-225)=-226×$ |

$-3$ and $-75$ | $-3+(-75)=-78×$ |

$-5$ and $-45$ | $-5+(-45)=-50×$ |

$-9$ and $-25$ | $-9+(-25)=-34×$ |

$-15$ and $-15$ | $-15+(-15)=-30✓$ |

$9x_{2}−$