# 1.4 Polynomials  (Page 4/15)

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$\begin{array}{ccc}\hfill \left(x+5\right)\left(x-5\right)& =& {x}^{2}-25\hfill \\ \hfill \left(x+11\right)\left(x-11\right)& =& {x}^{2}-121\hfill \\ \hfill \left(2x+3\right)\left(2x-3\right)& =& 4{x}^{2}-9\hfill \end{array}$

Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.

Is there a special form for the sum of squares?

No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.

## Difference of squares

When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term.

$\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$

Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares.

1. Square the first term of the binomials.
2. Square the last term of the binomials.
3. Subtract the square of the last term from the square of the first term.

## Multiplying binomials resulting in a difference of squares

Multiply $\text{\hspace{0.17em}}\left(9x+4\right)\left(9x-4\right).$

Square the first term to get $\text{\hspace{0.17em}}{\left(9x\right)}^{2}=81{x}^{2}.\text{\hspace{0.17em}}$ Square the last term to get $\text{\hspace{0.17em}}{4}^{2}=16.\text{\hspace{0.17em}}$ Subtract the square of the last term from the square of the first term to find the product of $\text{\hspace{0.17em}}81{x}^{2}-16.$

Multiply $\text{\hspace{0.17em}}\left(2x+7\right)\left(2x-7\right).$

$4{x}^{2}-49$

## Performing operations with polynomials of several variables

We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example:

## Multiplying polynomials containing several variables

Multiply $\text{\hspace{0.17em}}\left(x+4\right)\left(3x-2y+5\right).$

Follow the same steps that we used to multiply polynomials containing only one variable.

Multiply $\left(3x-1\right)\left(2x+7y-9\right).$

$\text{\hspace{0.17em}}6{x}^{2}+21xy-29x-7y+9$

Access these online resources for additional instruction and practice with polynomials.

## Key equations

 perfect square trinomial ${\left(x+a\right)}^{2}=\left(x+a\right)\left(x+a\right)={x}^{2}+2ax+{a}^{2}$ difference of squares $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$

## Key concepts

• A polynomial is a sum of terms each consisting of a variable raised to a non-negative integer power. The degree is the highest power of the variable that occurs in the polynomial. The leading term is the term containing the highest degree, and the leading coefficient is the coefficient of that term. See [link] .
• To multiply polynomials, use the distributive property to multiply each term in the first polynomial by each term in the second. Then add the products. See [link] .
• FOIL (First, Outer, Inner, Last) is a shortcut that can be used to multiply binomials. See [link] .
• Perfect square trinomials and difference of squares are special products. See [link] and [link] .
• Follow the same rules to work with polynomials containing several variables. See [link] .

## Verbal

Evaluate the following statement: The degree of a polynomial in standard form is the exponent of the leading term. Explain why the statement is true or false.

The statement is true. In standard form, the polynomial with the highest value exponent is placed first and is the leading term. The degree of a polynomial is the value of the highest exponent, which in standard form is also the exponent of the leading term.

Many times, multiplying two binomials with two variables results in a trinomial. This is not the case when there is a difference of two squares. Explain why the product in this case is also a binomial.

You can multiply polynomials with any number of terms and any number of variables using four basic steps over and over until you reach the expanded polynomial. What are the four steps?

Use the distributive property, multiply, combine like terms, and simplify.

State whether the following statement is true and explain why or why not: A trinomial is always a higher degree than a monomial.

## Algebraic

For the following exercises, identify the degree of the polynomial.

$7x-2{x}^{2}+13$

2

$14{m}^{3}+{m}^{2}-16m+8$

$-625{a}^{8}+16{b}^{4}$

8

$200p-30{p}^{2}m+40{m}^{3}$

${x}^{2}+4x+4$

2

$6{y}^{4}-{y}^{5}+3y-4$

For the following exercises, find the sum or difference.

$\left(12{x}^{2}+3x\right)-\left(8{x}^{2}-19\right)$

$4{x}^{2}+3x+19$

$\left(4{z}^{3}+8{z}^{2}-z\right)+\left(-2{z}^{2}+z+6\right)$

$\left(6{w}^{2}+24w+24\right)-\left(3w{}^{2}-6w+3\right)$

$3{w}^{2}+30w+21$

$\left(7{a}^{3}+6{a}^{2}-4a-13\right)+\left(-3{a}^{3}-4{a}^{2}+6a+17\right)$

$\left(11{b}^{4}-6{b}^{3}+18{b}^{2}-4b+8\right)-\left(3{b}^{3}+6{b}^{2}+3b\right)$

$11{b}^{4}-9{b}^{3}+12{b}^{2}-7b+8$

$\left(49{p}^{2}-25\right)+\left(16{p}^{4}-32{p}^{2}+16\right)$

For the following exercises, find the product.

$\left(4x+2\right)\left(6x-4\right)$

$24{x}^{2}-4x-8$

$\left(14{c}^{2}+4c\right)\left(2{c}^{2}-3c\right)$

$\left(6{b}^{2}-6\right)\left(4{b}^{2}-4\right)$

$24{b}^{4}-48{b}^{2}+24$

$\left(3d-5\right)\left(2d+9\right)$

$\left(9v-11\right)\left(11v-9\right)$

$99{v}^{2}-202v+99$

$\left(4{t}^{2}+7t\right)\left(-3{t}^{2}+4\right)$

$\left(8n-4\right)\left({n}^{2}+9\right)$

$8{n}^{3}-4{n}^{2}+72n-36$

For the following exercises, expand the binomial.

${\left(4x+5\right)}^{2}$

${\left(3y-7\right)}^{2}$

$9{y}^{2}-42y+49$

${\left(12-4x\right)}^{2}$

${\left(4p+9\right)}^{2}$

$16{p}^{2}+72p+81$

${\left(2m-3\right)}^{2}$

${\left(3y-6\right)}^{2}$

$9{y}^{2}-36y+36$

${\left(9b+1\right)}^{2}$

For the following exercises, multiply the binomials.

$\left(4c+1\right)\left(4c-1\right)$

$16{c}^{2}-1$

$\left(9a-4\right)\left(9a+4\right)$

$\left(15n-6\right)\left(15n+6\right)$

$225{n}^{2}-36$

$\left(25b+2\right)\left(25b-2\right)$

$\left(4+4m\right)\left(4-4m\right)$

$-16{m}^{2}+16$

$\left(14p+7\right)\left(14p-7\right)$

$\left(11q-10\right)\left(11q+10\right)$

$121{q}^{2}-100$

For the following exercises, multiply the polynomials.

$\left(2{x}^{2}+2x+1\right)\left(4x-1\right)$

$\left(4{t}^{2}+t-7\right)\left(4{t}^{2}-1\right)$

$16{t}^{4}+4{t}^{3}-32{t}^{2}-t+7$

$\left(x-1\right)\left({x}^{2}-2x+1\right)$

$\left(y-2\right)\left({y}^{2}-4y-9\right)$

${y}^{3}-6{y}^{2}-y+18$

$\left(6k-5\right)\left(6{k}^{2}+5k-1\right)$

$\left(3{p}^{2}+2p-10\right)\left(p-1\right)$

$3{p}^{3}-{p}^{2}-12p+10$

$\left(4m-13\right)\left(2{m}^{2}-7m+9\right)$

$\left(a+b\right)\left(a-b\right)$

${a}^{2}-{b}^{2}$

$\left(4x-6y\right)\left(6x-4y\right)$

${\left(4t-5u\right)}^{2}$

$16{t}^{2}-40tu+25{u}^{2}$

$\left(9m+4n-1\right)\left(2m+8\right)$

$\left(4t-x\right)\left(t-x+1\right)$

$4{t}^{2}+{x}^{2}+4t-5tx-x$

$\left({b}^{2}-1\right)\left({a}^{2}+2ab+{b}^{2}\right)$

$\left(4r-d\right)\left(6r+7d\right)$

$24{r}^{2}+22rd-7{d}^{2}$

$\left(x+y\right)\left({x}^{2}-xy+{y}^{2}\right)$

## Real-world applications

A developer wants to purchase a plot of land to build a house. The area of the plot can be described by the following expression: $\text{\hspace{0.17em}}\left(4x+1\right)\left(8x-3\right)\text{\hspace{0.17em}}$ where x is measured in meters. Multiply the binomials to find the area of the plot in standard form.

$32{x}^{2}-4x-3\text{\hspace{0.17em}}$ m 2

A prospective buyer wants to know how much grain a specific silo can hold. The area of the floor of the silo is $\text{\hspace{0.17em}}{\left(2x+9\right)}^{2}.\text{\hspace{0.17em}}$ The height of the silo is $\text{\hspace{0.17em}}10x+10,$ where x is measured in feet. Expand the square and multiply by the height to find the expression that shows how much grain the silo can hold.

## Extensions

For the following exercises, perform the given operations.

${\left(4t-7\right)}^{2}\left(2t+1\right)-\left(4{t}^{2}+2t+11\right)$

$32{t}^{3}-100{t}^{2}+40t+38$

$\left(3b+6\right)\left(3b-6\right)\left(9{b}^{2}-36\right)$

$\left({a}^{2}+4ac+4{c}^{2}\right)\left({a}^{2}-4{c}^{2}\right)$

${a}^{4}+4{a}^{3}c-16a{c}^{3}-16{c}^{4}$

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because x is 6
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if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
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