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2. Quadratic Functions
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Algebra 2

2.01 Factor polynomials

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Factor polynomials

Factoring a polynomial is a process of expressing a polynomial as a product of its factors. In other words, it is the inverse process of multiplying polynomials. A polynomial is considered completely factored when none of its factors can be written as a product of polynomials with a lower degree.

We have learned several different strategies, including a few identities, that can help us factor a polynomial.

  • Factoring using Greatest Common Factor (GCF):

    A greatest common factor from each term of a polynomial is factored out ax+ay+\ldots=a(x+y+\ldots)

  • Factoring by grouping:

    A method for factoring an expression containing at least four terms, by grouping the terms in pairs and taking out common factors ax + ay + bx + by = a\left(x + y\right) + b\left(x + y\right) = \left(x+y\right)\left(a+b\right)

  • Factoring quadratic trinomials:

    A trinomial that can be expressed as the product of two binomials ax^{2} + bx + c= \left(mx + p\right)\left(nx + q\right) where mn=a,pq=c and np+mq=b

  • Perfect square trinomials:

    A trinomial that is formed by multiplying a binomial by itself a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2} \text{ or } a^{2} - 2 a b + b^{2} = \left(a - b\right)^{2}

  • Difference of two squares:

    The result of a perfect square being subtracted from another perfect square a^{2} - b^{2} = \left(a+b\right)\left(a-b\right)

Examples

Example 1

Factor the polynomials using an appropriate method.

a

x^2-17x+60

Worked Solution
Create a strategy

First, we always check for a GCF. As x does not have a coefficient, there is no GCF here. Because this is a trinomial where the leading coefficient is 1, we want to factor by finding two numbers that add to -17 and multiply to 60.

Apply the idea

Since we know we want it multiply to positive number and add to a negative number, we know both factors will be negative.

\text{Factors of }60-1,-60-2, -30-3, -20-4, -15-5, -12-6, -10
\text{Sum}-61-32-23-19-17-16

We need to choose factors of 60 that have sum of -17. Looking at our table above, we see the factors are -5 and -12.

\displaystyle x^2-17x+60\displaystyle =\displaystyle \left(x-5\right)\left(x-12\right)
Reflect and check

We can check that we factored correctly by multiplying the binomials and confirming we get original expression.

\displaystyle \left(x-5\right)\left(x-12\right)\displaystyle =\displaystyle x^2-12x-5x+60Multiply using the distributive property
\displaystyle {}\displaystyle =\displaystyle x^2-17x+60Combine like terms

We can confirm our expression is the same as original which shows we factored correctly.

b

8 x y^{4} - 16 x^{3} y^{2} + 4 x y^{3} .

Worked Solution
Create a strategy

We will first find the GCF of 8 x y^{4} , - 16 x^{3} y^{2} , and 4 x y^{3} . We can then rewrite each expression as a product of the GCF and any remaining factors.

Apply the idea

The GCF of 8 x y^{4} , - 16 x^{3} y^{2} \text{ and } 4 x y^{3} is 4xy^{2} . So we will divide each term by 4xy^{2} :

\displaystyle 8 x y^{4} - 16 x^{3} y^{2} + 4 x y^{3} \displaystyle =\displaystyle 4xy^{2} \left(\dfrac{8 x y^{4} }{4xy^{2} } - \dfrac{16 x^{3} y^{2} }{4xy^{2} } + \dfrac{4 x y^{3} }{4xy^{2} }\right)
\displaystyle =\displaystyle 4xy^{2} \left(\dfrac{2 x y^{4} }{xy^{2} } - \dfrac{4 x^{3} y^{2} }{xy^{2} } + \dfrac{ x y^{3} }{xy^{2} }\right)Divide the coefficients
\displaystyle =\displaystyle 4xy^{2} \left(2y^{2} - 4 x^{2} + y \right)Quotient of powers

The remaining terms do not have any other common factors, and the linear term does not contain both variables. Therefore, this expression is fully factored.

Reflect and check

We can check the answer by distributing the multiplication and using the product of powers rule for exponents:

\displaystyle 4xy^{2} \left(2y^{2} - 4 x^{2} + yz^{4} \right)\displaystyle =\displaystyle 4xy^{2} \left(2y^{2}\right) - 4xy^{2} \left(4 x^{2}\right) + 4xy^{2} \left(y\right)
\displaystyle =\displaystyle 8 x y^{4} - 16 x^{3} y^{2} + 4 x y^{3}

Notice that, if no mistakes have been made, these are the same steps, just in reverse.

c

121m^{2}-64

Worked Solution
Create a strategy

The expression has 2 terms that are both perfect squares, and the terms are subtracted. This means we can factor using the difference of two squares identity: A^{2}-B^{2}=\left(A+B\right) \left(A-B\right)

Apply the idea

Since (11m)^2=121m^2 and 8^2=64, we can use A^{2}-B^{2}=\left(A+B\right) \left(A-B\right), where A=11m and B=8.

\displaystyle 121m^{2}-64\displaystyle =\displaystyle \left(11m+8\right)\left(11m-8\right)
d

x^2-12xy+36y^2

Worked Solution
Create a strategy

Notice that there are no common factors, and the first and last terms are perfect squares. Let's check if this expression meets the criteria for a perfect square trinomial: \left(A-B\right)^2=A^2-2AB+B^2 We can see that A^2=x^2, which means A=x. Next, we see that B^2=36y^2, so B=6y. Now, let's check whether -2AB=-12xy: -2\cdot x\cdot 6y=-12xy

This means the expression is a perfect square trinomial, so we can use that identity to factor the expression.

Apply the idea

Using the perfect square trinomial identity \left(A-B\right)^2=A^2-2AB+B^2 with A=x and B=6y:

\displaystyle x^2-12x+36y^2\displaystyle =\displaystyle \left(x-6y\right)^2
Reflect and check

We can check the answer by expanding it again. This can also help us notice patterns when factoring expressions with two variables.

\displaystyle \left(x-6y\right)^2\displaystyle =\displaystyle (x-6y)(x-6y)Expand
\displaystyle =\displaystyle x^2-6xy-6xy+36y^2Multiply using the distributive property
\displaystyle =\displaystyle x^2-12xy+36y^2Combine like terms
e

18m^{3}-2mn^2+9m^2n-n^3

Worked Solution
Create a strategy

First, we want to see if all terms share a common factor. These terms do not, so we can check for factoring by grouping next because the expression has 4 terms.

Apply the idea
\displaystyle 18m^{3}-2mn^2+9m^2n-n^3\displaystyle =\displaystyle \left(18m^{3}-2mn^2\right)+\left(9m^2n-n^3\right)Group based on common factors
\displaystyle =\displaystyle 2m\left(9m^{2}-n^{2}\right) + n\left(9m^{2}-n^{2}\right)Factor out each GCF (2m and n)
\displaystyle =\displaystyle \left(2m+n\right)\left(9m^{2}-n^{2}\right)Factor out the common binomial factor

Observe that 9m^{2}-n^{2}=\left(3m\right)^{2}-\left(n\right)^{2}. This means that 9m^{2}-n^{2} is a difference of two squares, so we use the formula in factoring:

\displaystyle a^{2}-b^{2}\displaystyle =\displaystyle (a-b)(a+b)Formula of difference of two squares
\displaystyle \left(3m\right)^{2}-\left(n\right)^{2}\displaystyle =\displaystyle (3m-n)(3m+n)Substitue a=3m and b=n

If we substitute 9m^{2}-n^{2}=(3m-n)(3m+n), we get 18m^{3}-2mn^2+9m^2n-n^3=\left(2m+n\right)(3m-n)(3m+n)

Reflect and check

Alternatively, we can group 18m^{3} \text{ and } 9m^{2}n and -2mn^{2} \text{ and } -n^{3} together and get the same answer.

\displaystyle 18m^{3}-2mn^2+9m^2n-n^3\displaystyle =\displaystyle \left(18m^{3}+9m^2n \right)+ \left(-2mn^2-n^3\right)Group based on common factors
\displaystyle =\displaystyle 9m^{2}\left(2m+n\right) + \left(-n^{2}\right)\left(2m+n\right)Factor out each GCF (9m^2 and -n^2)
\displaystyle =\displaystyle \left(9m^{2} - n^{2}\right)\left(2m + n\right)Factor out the common binomial factor
\displaystyle =\displaystyle \left(3m +n\right)\left(3m -n\right)\left(2m + n\right)Apply the formula for difference of two squares
f

6 x^{4} - 10x^{3} - 24x^{2}

Worked Solution
Create a strategy

First, we look for the greatest common factor for all three terms and factor it out. Then we look at the simpler trinomial to see if it factors further. If it does, we find the values of r and s that multiply to ac and add to b. After finding those values, we write the trinomial in the form ax^{2} + rx + sx + c and factor it by grouping.

Apply the idea

The greatest common factor of the three terms is 2x^{2}. We factor out 2x^{2} and get 2x^{2}\left(3 x^{2} - 5x - 12\right)

For 3 x^{2} - 5x - 12, we need to find two numbers that have a product of ac = \left(3\right)\left(-12\right) = -36 and sum of b = -5.

The factors of -36 include \pm1, \pm2, \pm3,\pm4,\pm6,\pm9,\pm18 and \pm36. Among these factors, 4 and -9 are the only numbers that add to -5 and multiply to -36. We can let r=4 and s=-9. Next, we write 3 x^{2} - 5x - 12 in the form ax^{2} + rx + sx + c. Substituting the values for r and s, we get

3x^{2} +4x -9 x - 12

Now, we factor this expression by grouping

\displaystyle 3x^{2} +4x -9x -12\displaystyle =\displaystyle (3x^{2} +4x) +(-9x -12)Group based on common factors
\displaystyle =\displaystyle x\left(3x + 4\right) - 3\left(3x + 4\right)Factor out each GCF (x and -3)
\displaystyle =\displaystyle \left(x-3\right)\left(3x+4\right)Factor out the common binomial factor

Therefore, 6 x^{4} - 10x^{3} - 24x^{2}=2x^{2}\left(x-3\right)\left(3x+4\right).

Reflect and check

We can check the answer by multiplying the factored form 2x^{2}\left(x-3\right)\left(3x+4\right).

\displaystyle 2x^{2}\left(x-3\right)\left(3x+4\right)\displaystyle =\displaystyle 2x^{2}\left(x\left(3x+4\right)-3\left(3x+4\right)\right)Distributive property
\displaystyle =\displaystyle 2x^{2}\left(3x^{2} +4x -9x - 12\right)Distributive property
\displaystyle =\displaystyle 6 x^{4} + 8x^{3} -18x^{3} - 24x^{2}Distributive property
\displaystyle =\displaystyle 6 x^{4} - 10x^{3} - 24x^{2}Combine like terms
Idea summary

The factoring methods and identities we can use to fully factor polynomials are

  • Factoring using the GCF
  • Factoring by grouping
  • Factoring quadratic trinomials
  • Perfect square trinomial identity
  • Difference of two squares identity

Outcomes

A2.EO.3

The student will perform operations on polynomial expressions in two or more variables and factor polynomial expressions in one and two variables.

A2.EO.3b

Factor polynomials completely in one and two variables with no more than four terms over the set of integers.

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