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9.9: Factor Polynomials Using Special Products

Difficulty Level: Basic Created by: CK-12
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Suppose the area of a square playground is 10,000 square feet. Instead of taking the square root of the area to find the length of one of the playground's sides, you can set up the equation \begin{align*}s^2 = 10,000\end{align*}s2=10,000, subtract 10,000 from both sides to get \begin{align*}s^2 - 10,000 = 0\end{align*}s210,000=0, and solve for \begin{align*}s\end{align*}s by factoring \begin{align*}s^2 -10,000\end{align*}s210,000 on the left side of the equation. But how would you factor this? 

Factoring Polynomials Using Special Products

When we learned how to multiply binomials, we talked about two special products: the Sum and Difference Formula and the Square of a Binomial Formula. Now, we will learn how to recognize and factor these special products.

Factoring the Difference of Two Squares

We use the Sum and Difference Formula to factor the difference of two squares. A difference of two squares is a quadratic polynomial in this form: \begin{align*}a^2-b^2\end{align*}a2b2. Both terms in the polynomial are perfect squares. In a case like this, the polynomial factors into the sum and difference of the square root of each term.

\begin{align*}a^2-b^2=(a+b)(a-b)\end{align*}a2b2=(a+b)(ab)

In these problems, the key is figuring out what the \begin{align*}a\end{align*}a and \begin{align*}b\end{align*}b terms are. 

Let's factor the following binomials:

  1. \begin{align*}x^2-9\end{align*}x29

First, rewrite \begin{align*}x^2-9\end{align*}x29 as \begin{align*}x^2-3^2\end{align*}x232. Now it is obvious that it is a difference of squares.

We substitute the values of \begin{align*}a\end{align*}a and \begin{align*}b\end{align*}b in the Sum and Difference Formula:

\begin{align*}(x+3)(x-3)\end{align*}

The answer is \begin{align*}x^2-9=(x+3)(x-3)\end{align*}.

  1. \begin{align*}x^2y^2-1\end{align*}

First, rewrite \begin{align*}x^2y^2-1\end{align*} as \begin{align*}(xy)^2-1^2\end{align*}.

This factors as \begin{align*}(xy+1)(xy-1)\end{align*}.

Factoring Perfect Square Trinomials

A perfect square trinomial has the form:

\begin{align*}a^2+2ab+b^2 \qquad \text{or} \qquad a^2-2ab+b^2\end{align*}

The factored form of a perfect square trinomial has the form:

\begin{align*}(a+b)^2 \ &if \ a^2+2(ab)+b^2\\ &\text{and}\\ (a-b)^2 \ &if \ a^2-2(ab)+b^2\end{align*}

In these problems, the key is figuring out what the \begin{align*}a\end{align*} and \begin{align*}b\end{align*} terms are. 

Let's factor the following trinomials:

  1. \begin{align*}x^2+8x+16\end{align*}

Check that the first term and the last term are perfect squares.

\begin{align*}x^2+8x+16 \qquad \text{as} \qquad x^2+8x+4^2.\end{align*}

Check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them.

\begin{align*}x^2+8x+16 \qquad \text{as} \qquad x^2+2 \cdot 4 \cdot x+4^2\end{align*}

This means we can factor \begin{align*}x^2+8x+16\end{align*} as \begin{align*}(x+4)^2\end{align*}.

  1. \begin{align*}x^2-4x+4\end{align*}

Rewrite \begin{align*}x^2-4x+4\end{align*} as \begin{align*}x^2+2 \cdot (-2) \cdot x+(-2)^2\end{align*}.

We notice that this is a perfect square trinomial and we can factor it as \begin{align*}(x-2)^2\end{align*}.

Solving Polynomial Equations Involving Special Products

We have learned how to factor quadratic polynomials that are helpful in solving polynomial equations like \begin{align*}ax^2+bx+c=0\end{align*}. Remember that to solve polynomials in expanded form, we use the following steps:

Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0.

Step 2: Factor the polynomial completely.

Step 3: Use the Zero Product Property to set each factor equal to zero.

Step 4: Solve each equation from step 3.

Step 5: Check your answers by substituting your solutions into the original equation.

Let's solve the following equation:

\begin{align*}x^2+4x+4=0\end{align*}

Rewrite: No need to rewrite because it is already in the correct form.

Factor: Note that the first and last term are perfect squares. Also note that the middle term is twice the product of the square roots of the first and last terms. The left side of the equation is a perfect square trinomial. We can use the formula to factor it:

\begin{align*}x^2 + 4x + 4 &= 0\\ (x + 2)^2 &=0\end{align*} 

Set each factor equal to zero: Since there is only one distinct factor, we only need to set \begin{align*}x+2\end{align*} equal to 0.  

\begin{align*}x+2=0\end{align*}

Solve:

\begin{align*}x=-2\end{align*}

Check: Substitute the solution back into the original equation.

\begin{align*}(-2)^2+4(-2)+4&=4+(-8)+4=0\\ \end{align*}

The solution is -2. 

   

 

Examples

Example 1

Earlier, you were told that the area of a square playground is 10,000 square feet. Instead of taking the square root of the area to find the length of one of the playground's sides, you can set up the equation \begin{align*}s^2 = 10,000\end{align*}, subtract 10,000 from both sides to get \begin{align*}s^2 - 10,000 = 0\end{align*}, and solve for \begin{align*}s\end{align*} by factoring \begin{align*}s^2 -10,000\end{align*} on the left side of the equation. How would you factor this?

Using the steps to solve an equation, we can solve for \begin{align*}s\end{align*} in the equation \begin{align*}s^2 = 10,000\end{align*}.

Rewrite: Subtract 10,000 from each side to put the equation in standard form.

\begin{align*}s^2-10,000 = 10,000-10,000\\ s^2-10,000 = 0\end{align*} 

Factor: Note that the left side of the equation is the difference of two perfect squares. We can use the formula to factor it:

\begin{align*}s^2 -10,000= 0\\ (s -100)(s+100)&=0\end{align*} 

Set each factor equal to zero: 

 

\begin{align*}s+100 = 0 \text{ and } s-100=0\end{align*}

Solve:

\begin{align*}s=-100 \text{ and } s=100\end{align*}

Check: Substitute each of the solutions back into the original equation.

\begin{align*}100^2 = 225 \text{ and } (-100)^2 = 225 \end{align*}

The solutions are 100 and -100. However, \begin{align*}s\end{align*} represents the length of a side of a playground. The length of an object cannot be negative so the only answer is 100. 

Example 2

Solve the equation \begin{align*}x^2 =225\end{align*}

Note that this is very similar to Example 1. 

Rewrite: Subtract 225 from each side to put the equation in standard form.

\begin{align*}x^2-225 = 225-225\\ x^2-225 = 0\end{align*} 

Factor: Note that the left side of the equation is the difference of two perfect squares. We can use the formula to factor it:

\begin{align*}x^2 -225= 0\\ (x -15)(x+15)&=0\end{align*} 

Set each factor equal to zero: 

 

\begin{align*}x+15 = 0 \text{ and } x-15=0\end{align*}

Solve:

\begin{align*}x=-15 \text{ and } x=15\end{align*}

Check: Substitute each of the solutions back into the original equation.

\begin{align*}15^2 = 225 \text{ and } (-15)^2 = 225 \end{align*}

The solutions are 15 and -15.

Review

Factor the following perfect square trinomials.

  1. \begin{align*}x^2+8x+16\end{align*}
  2. \begin{align*}x^2-18x+81\end{align*}
  3. \begin{align*}-x^2+24x-144\end{align*}
  4. \begin{align*}x^2+14x+49\end{align*}
  5. \begin{align*}4x^2-4x+1\end{align*}
  6. \begin{align*}25x^2+60x+36\end{align*}
  7. \begin{align*}4x^2-12xy+9y^2\end{align*}
  8. \begin{align*}x^4+22x^2+121\end{align*}

Factor the following differences of squares.

  1. \begin{align*}x^2-4\end{align*}
  2. \begin{align*}x^2-36\end{align*}
  3. \begin{align*}-x^2+100\end{align*}
  4. \begin{align*}x^2-400\end{align*}
  5. \begin{align*}9x^2-4\end{align*}
  6. \begin{align*}25x^2-49\end{align*}
  7. \begin{align*}-36x^2+25\end{align*}
  8. \begin{align*}16x^2-81y^2\end{align*}

Solve the following quadratic equations using factoring.

  1. \begin{align*}x^2-11x+30=0\end{align*}
  2. \begin{align*}x^2+4x=21\end{align*}
  3. \begin{align*}x^2+49=14x\end{align*}
  4. \begin{align*}x^2-64=0\end{align*}
  5. \begin{align*}x^2-24x+144=0\end{align*}
  6. \begin{align*}4x^2-25=0\end{align*}
  7. \begin{align*}x^2+26x=-169\end{align*}
  8. \begin{align*}-x^2-16x-60=0\end{align*}

Mixed Review

  1. Find the value for \begin{align*}k\end{align*} that creates an infinite number of solutions to the system \begin{align*}\begin{cases} 3x+7y=1\\ kx-14y=-2 \end{cases}\end{align*}.
  2. A restaurant has two kinds of rice, three choices of mein, and four kinds of sauce. How many plate combinations can be created if you choose one of each?
  3. Graph \begin{align*}y-5= \frac{1}{3}(x+4)\end{align*}. Identify its slope.
  4. $600 was deposited into an account earning 8% interest compounded annually.
    1. Write the exponential model to represent this situation.
    2. How much money will be in the account after six years?
  1. Divide: \begin{align*}4 \frac{8}{9} \div -3\frac{1}{5}\end{align*}.
  2. Identify an integer than is even and not a natural number.

Review (Answers)

To see the Review answers, open this PDF file and look for section 9.9. 

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Vocabulary

difference of two squares

The difference of two squares has the form a^2-b^2=(a+b)(a-b).

Difference of Squares

A difference of squares is a quadratic equation in the form a^2-b^2.

Perfect Square Trinomial

A perfect square trinomial is a quadratic expression of the form a^2+2ab+b^2 (which can be rewritten as (a+b)^2) or a^2-2ab+b^2 (which can be rewritten as (a-b)^2).

Quadratic form

A polynomial in quadratic form looks like a trinomial or binomial and can be factored like a quadratic expression.

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Difficulty Level:
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Date Created:
Feb 24, 2012
Last Modified:
Aug 16, 2016
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