Factoring Polynomials

Factor the value of 20.

Solution

There are many ways to break down the product of 20.

Take 2 x 10 = 20, for instance.

However, 10 is not a prime and so we can further break down this product, as below

2 x 2 x 5 = 20.

Now, we can no longer simplify the components that make up this product. Thus, it has been completely factorised.

Factor out the GCF from the following expression,

$15x^4-20x^3+35x^2$

Solution

Here, we can factor out 5 and x² from every term. In doing so, we obtain

$15x^4-20x^3+35x^2=5x^2(3x^2-4x+7)$

Factor out the GCF from the following expression,

$x^3{y}^2+3x^{4}y+5x^5{y}^3$

Solution

Notice that each term contains both x³and y. Thus, factoring these out yields

$x^3{y}^2+3x^{4}y+5x^5{y}^3=x^{3}y(y+3x+5x^2{y}^2)$

Factor out the GCF from the following expression,

$9x^2(2x+7)-12x(2x+7)$

Solution

Firstly, observe that each term contains the binomial (2x + 7). Factoring this out yields

$9x^2(2x+7)-12x(2x+7)=(2x+7)(9x^2-12x)$

Now notice that we can further take out 3x from (9x² - 12x). This will give us the final factorized form as

$(2x+7)(9x^2-12x)=3x(2x+7)(3x-4)$

Factorize the quadratic trinomial below,

$x^2+x-12$

Solution

Steps 1 and 2: We start by looking at the first term, x². Factoring this may look like the expression below

$x^2+x-12=(x+\square )(x+\square )$

since multiplying x by x gives us x².

Step 3: To determine the values that go into the blank spaces, we must find a pair of numbers that multiply to get –12 and add to get 1. We do this by looking for factors of the last term, –12. Listing this down, we get the following pairs.

(-1)(12), (1)(-12), (-2)(6), (2)(-6), (-3)(4), (3)(-4)

Step 4: We find that the pair (-3)(4) satisfies the polynomial.

Step 5: Writing this in the final factored form, we obtain

$x^2+x-12=(x-3)(x+4)$

Factorize the quadratic trinomial below,

$6x^2+11x-10$

Solution

Steps 1 and 2: Looking at the first term, we have two possible factored forms,

$$\begin{gathered} 6x^2+11x-10=(3x+\square )(2x+\square ) \\ and \\ 6x^2+11x-10=(6x+\square )(x+\square ) \end{gathered}$$

Step 5: With some trial and error, we obtain the final factorized form as

$6x^2+11x-10=(3x-2)(2x+5)$

Factorize the quadratic trinomial below,

$3x^2+2x-8$

Solution

$3x^2+2x-8=(3x+\square )(x+\square )$

Step 3: We must find a pair of numbers that multiply to get –8 and add up to get 2. Looking at the final term, –8, we can list down its factors as

(-1)(8), (1)(-8), (-2)(4), (2)(-4)

Step 4: Plugging these into the blank spaces one by one tells us that (-4)(2) is the correct pair.

Step 5: In doing so, our final factorized form becomes

$3x^2+2x-8=(3x-4)(x+2)$

Use the grouping method to factorize the polynomial $x^3+2x^2-3x-6$

Solution

Step 1: We start by grouping the first two terms and last two terms as below.

$(x^3+2x^2)+(-3x-6)$

Step 2: Notice that we can factor out x² from the first group and –3 from the second group.

$x^2(x+2)+(-3)(x+2)$

Step 3: Now observing that both groups have a common binomial (x + 2), we can factor this out to obtain

$x^3+2x^2-3x-6=(x+2)(x^2-3)$

Use the grouping method to factorize the polynomial $3xy-21y+5x-35$

Solution

Step 1: As before, we start by grouping the first two terms and last two terms as below.

$(3xy-21y)+(5x-35)$

Step 2: Notice that we can factor out 3y from the first group and 5 from the second group.

$3y(x-7)+5(x-7)$

Step 3: Now observing that both groups have a common binomial (x – 7), we can factor this out to obtain

$3xy-21y+5x-35=(x-7)(3y+5)$

Use the grouping method to factorize the polynomial $x^5+x-2x^4-2$

Solution

Step 1: Start by grouping the first two terms and last two terms as below.

$(x^5+x)+(-2x^4-2)$

Step 2: Notice that we can factor out x from the first group and –2 from the second group.

$x(x^4+1)+(-2)(x^4+1)$

Step 3: Now observing that both groups have a common binomial (x⁴ + 1), we can factor this out to obtain

$x^5+x-2x^4-2=(x^4+1)(x-2)$

Use the grouping method to solve the trinomial, $x^2-8x-9$

Solution

Here, a = 1, b = –8 and c = –9. The product of ac is –9. We need a pair of numbers that add up to –8 and yield –9 when multiplied together. Our 4-term polynomial should look like this:

$x^2+\square x+\square x-9$

To find the missing numbers, we shall look at the middle term, –8. Notice that if we add 1 to –9 we obtain –8. The product of 1 and –9 is –9, that is ac. We thus have the following 4-term polynomial.

$x^2+x-9x-9$

Now, let us group the first two terms and the last two terms as below.

$(x^2+x)+(-9x-9)$

Notice that we can factor out x from the first group and –9 from the second group.

$x(x+1)+(-9)(x+1)$

Now observing that both groups have a common binomial (x + 1), we can factor this out to obtain

$x^2-8x-9=(x+1)(x-9)$

Completely factorize the expression $3x^5-7x^4-26x^3$

Solution

First, notice that we can factor out x³ from each term.

$x^3(3x^2-7x-26)$

Now we have the quadratic trinomial $3x^2-7x-26$. From the first term the factorized form will take on the structure.

$3x^5-7x^4-26x^3=x^3(3x+\square )(x+\square )$

We need to find a pair of numbers that multiply to get –26 and add to get –7. Let us look at the factors of the term –26.

(-1)(26), (1)(-26), (-2)(13), (2)(-13)

Using trial and error, we find that the pair (–13)(2) satisfies the criteria for this factored form, and so the fully factorised form for this polynomial is

$3x^5-7x^4-26x^3=x^3(3x-13)(x+2)$

Fully factorize the expression $x^3-2x^2-9x+18$

Solution

We begin by using the grouping method to factorize this. Grouping the first two terms and last two terms, we obtain

$(x^3-2x^2)+(-9x+18)$

We can factor out x² from the first group and –9 from the second group.

$x^2(x-2)+(-9)(x-2)$

We now have the common factor of (x – 2) in both groups. Factoring this out yields,

$x^3-2x^2-9x+18=(x-2)(x^2-9)$

Now, you may think that we have completed factoring our expression here. However, this is not the case. Observe the binomial $(x^2-9)$. This is an example of a perfect square binomial. We will look at this in more detail on the topic of Special Products.

To avoid confusion, we shall treat this as a quadratic trinomial where the middle term is essentially zero. By the first term, we can see that the factorized form must look like the following

$x^3-2x^2-9x+18=(x-2)(x+\square )(x+\square )$

Now we need to look for a pair of numbers that multiply to get –9 and add up to get 0. The pair (–3)(3) is the most possible answer here. Thus, we obtain

$x^3-2x^2-9x+18=(x-2)(x-3)(x+3)$

Factorize the expression $x^4-13x^2+36$

Solution

First, notice that this expression looks similar to a quadratic trinomial. Let us use grouping to solve this. This 3-term polynomial will take the form of the 4-term polynomial below

$x^4+\square x^2+\square x^2+36$

We must now find 2 numbers that add up to get –13 and multiply to get 36. By trial and error, we find that the pair –4 and –9 satisfy this criterion.

$x^4-4x^2-9x^2+36$

Grouping the first two terms and last two terms, we obtain

$(x^4-4x^2)+(-9x^2+36)$

Factoring x² from the first group and –9 from the second group, we get

$x^2(x^2-4)+(-9)(x^2-4)$

Notice that we can factor out (x² - 4) from both groups, yielding

$x^4-13x^2+36=(x^2-4)(x^2-9)$

But we are not done yet! Just like the previous examples, we have the perfect square binomials $(x^2-4)and(x^2-9)$. Thus, we can further factorize this. Using the same principle as before, we finally obtain

$x^4-13x^2+36=(x-2)(x+2)(x-3)(x+3)$

Find the solutions to the polynomial below using the Zero Product Property

$3x^5-7x^4-26x^3=0$

Solution

From our solution above, the factorized form is

$x^3(3x-13)(x+2)=0$

As we have a product of 3 factors, we must have 3 solutions. Applying the Zero Product Property, we obtain

$$\begin{gathered} x^3=0 \\ 3x-13=0 \\ x+2=0 \end{gathered}$$

Solving these for x, we obtain 3 real roots

$x=-2,x=0andx=\frac{13}{3}$

Find the solutions to the polynomial below using the Zero Product Property

$x^3-2x^2-9x+18=0$

Solution

The factorized form is

$(x-2)(x-3)(x+3)=0$

As we have a product of 3 factors, we must have 3 solutions. Applying the Zero Product Property, we obtain

$$\begin{gathered} x-2=0 \\ x-3=0 \\ x+3=0 \end{gathered}$$

Solving these for x, we obtain 3 real roots

$x=-3,x=2andx=3$

The graph is shown below.

Example 1, Aishah Amri - StudySmarter Originals

Find the solutions to the polynomial below using the Zero Product Property

$x^4-13x^2+36=0$

Solution

We have the factorized form as

$(x-2)(x+2)(x-3)(x+3)=0$

As we have a product of 4 factors, we must have 4 solutions. Applying the Zero Product Property, we obtain

$$\begin{gathered} x-2=0 \\ x+2=0 \\ x-3=0 \\ x+3=0 \end{gathered}$$

Solving these for x, we obtain 4 real roots

$x=-3,x=-2,x=2andx=3$

The graph is shown below.

Example 2(1), Aishah Amri - StudySmarter Originals

Zooming into the x-intercepts, we have

Example 2(2), Aishah Amri - StudySmarter Originals

Solve ${(x-7)}^2=0$

Solution

Square root both sides.

$$\begin{gathered} \sqrt{{(x-7)}^2}=\sqrt{0} \\ \Rightarrow x-7=0 \end{gathered}$$

The square root of zero is zero. Solving this yields

$x=7$

Thus, we only have one real solution, as stated. The graph is shown below.

Example 3, Aishah Amri - StudySmarter Originals