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Reference

Method

When all terms in an expression contain a common factor, the expression can be rewritten as the product of such common factor and another factor that is the sum of the terms divided by the common factor. Consider for example the following expression.
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Factoring Binomials

$4x_{3}+6x_{2}−12x $

Notice that each term contains $x,$ thus factoring out $x$ would help in the factorization. However, factoring the greatest common factor, GCF, of the expression is preferred. In this case, the expression can be factored following the next three steps.
1

Find the GCF of the Expression

Start by finding the GCF of the expression. To do this, rewrite each term as the products of its factors.

$4x_{3}6x_{2}12x =2⋅2⋅x⋅x⋅x=2⋅3⋅x⋅x=2⋅2⋅3⋅x $

The GCF of the initial expression is $2x.$ 2

Rewrite Each Term in Terms of the GCF

Next, rewrite each term of the initial expression as the product of the GCF and another factor.

$4x_{3}6x_{2}12x =2x⋅2x_{2}=2x⋅3x=2x⋅6 $

One way of finding the corresponding factors is dividing the original terms by the GCF.
$2x4x_{3} 2x6x_{2} 2x12x =2x_{2}=3x=6 $

Therefore, the initial expression can be written as follows.
$4x_{3}+6x_{2}−12x⇕2x⋅2x_{2}+2x⋅3x−2x⋅6 $

3

Factor Out the GCF

Finally, using the Distributive Property, factor out the GCF.

$4x_{3}+6x_{2}−12x⇕2x(2x_{2}+3x−6) $

After this, the expression between the parentheses has to be studied to continue with the factorization, if possible. In the following applet, different binomials are factored by GCF.

Method

The product of a conjugate pair of binomials results in a difference of two squares. Using this relationship, the difference of two squares can be factored as the product of the sum and difference of two quantities.
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Justification of Factorization

$a_{2}−b_{2}=(a+b)(a−b) $

As an example, the following expression will be factored.
$9x_{2}−121 $

There are two steps to factor the expression as a difference of two squares.
1

Examine the Terms of the Expression

To factor an expression as a difference of two squares, the terms of the expression should be perfect squares.

$9x_{2}−121⇓(3x)_{2}−(11)_{2} $

As it can be seen, the terms of the expression are perfect squares. 2

Write as a Product of a Conjugate Pair of Binomials

The factored form of the expression can be expanded to justify the factorization.

Method

For a trinomial to be factorable as a perfect square trinomial, the first and last terms must be perfect squares and the middle term must be two times the square roots of the first and last terms. Consider the following expression.
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$16x_{2}+88x+121 $

To factor this trinomial, there are three steps.
1

Confirm That the First and Last Terms Are Perfect Squares

One good way to recognize if a trinomial is a perfect square trinomial is to look at its first and last terms. If they are both perfect squares, there is a good chance that it is a perfect square trinomial. In the given expression, the first and last terms can be written as the squares of $4x$ and $11,$ respectively.

$16x_{2}+88x+121⇓(4x)_{2}+88x+(11)_{2} $

These perfect squares show that the expression could be a perfect square trinomial. However, this is not enough to decide yet. 2

Confirm That the Middle Term Is Twice the Product of the Square Roots of First and Last Terms

The next step is to check whether the middle term is two times the square roots of the first and last terms.

$(4x)_{2}+88x+(11)_{2}⇓(4x)_{2}+2(4x)(11)+(11)_{2} $

It can be seen that the given expression satisfies this condition as well. 3

Write as a Square of a Binomial

Since the expression satisfies both conditions, it is a perfect square trinomial. Therefore, it can be written as a square of a binomial where $4x$ and $11$ are the first and second terms of the binomial, respectively.

$16x_{2}+88x+121=(4x+11)_{2} $

Method

To factor quadratic trinomials in the form $x_{2}+bx+c,$ two numbers $p$ and $q$ with a sum of $b$ and a product of $c$ must be found so that the trinomial is written as the product of $(x+p)$ and $(x+q).$
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$x_{2}+bx+c=(x+p)(x+q) $

As an example, the trinomial below will be factored.
$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) $

Method

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) $

Method

When a polynomial has four or more terms, they usually do not have a common factor. In some of these cases, it might be convenient to group the terms in pairs and factor the greatest common factor of each pair. Once this is done, another common factor might appear, allowing the polynomial to be factored. For example, consider the following polynomial.
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Rewriting the Terms First

$y=2x_{3}−x_{2}+6x−3 $

The terms of this polynomial share no common factor. However, the polynomial can be factored following the next three steps.
1

Group the Terms in Pairs

Start by grouping the first two terms and the last two terms.

$y=2x_{3}−x_{2}+6x−3⇕y=(2x_{3}−x_{2})+(6x−3) $

2

Factor Out the GCF of Each Pair

In the first pair, factor out the GCF, which is $x_{2}.$

$y=(2x_{3}−x_{2})+(6x−3)⇕y=x_{2}(2x−1)+(6x−3) $

For the second pair, the GCF is $3.$
$y=x_{2}(2x−1)+(6x−3)⇕y=x_{2}(2x−1)+3(2x−1) $

3

Factor Out the GCF of the Resulting Expression

If the polynomial is factorable, the second step should lead to a sum of two terms with a common factor.

$y=x_{2}(2x−1)+3(2x−1) $

In this case, the GCF of the terms is $(2x−1).$ Thus, factoring it out allows factoring the initial polynomial.
$y=x_{2}(2x−1)+3(2x−1)⇕y=(2x−1)(x_{2}+3) $

There might be polynomials that require some rewrites before they can be factored by grouping. If the rewrite is not done, the method could not work even though the polynomial is factorable. For example, consider the following polynomial.

$y=x_{3}−3x_{2}+4x−2 $

Grouping the terms in pairs and factoring out their corresponding GCF will not help in factoring the polynomial.
$y=(x_{3}−3x_{2})+(4x−2)⇓y=x_{2}(x−3)+2(2x−1) $

As seen, the terms between parentheses are not equal and therefore, they cannot be factored out. However, rewriting $-3x_{2}$ as $-x_{2}−2x_{2}$ and $4x$ as $2x+2x$ could be useful.
$y=x_{3}−3x_{2}+4x−2⇓y=x_{3}-3x_{2}−x_{2}−2x_{2} 4x+2x+2x −2 $

Now, there are six terms in the polynomial and they can be grouped in pairs.
$y=(x_{3}−x_{2})+(-2x_{2}+2x)+(2x−2) $

Next, factor out the GCF of each pair. In this case, factor out $x_{2},$ $-2x,$ and $2$ from the first, second, and third pair respectively.
$y=x_{2}(x−1)−2x(x−1)+2(x−1) $

Notice that $(x−1)$ is a common factor for the three terms. Therefore, it can be factored out.
$y=(x−1)(x_{2}−2x+2) $

This way the initial polynomial was written as the product of two factors. To complete the factorization, the quadratic polynomial needs to be factored. Note that the rewrites made here are particular for the given polynomial.
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