# 1.5 Factoring polynomials  (Page 3/6)

 Page 3 / 6

Factor a. $\text{\hspace{0.17em}}2{x}^{2}+9x+9\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}6{x}^{2}+x-1$

a. $\text{\hspace{0.17em}}\left(2x+3\right)\left(x+3\right)\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}\left(3x-1\right)\left(2x+1\right)$

## Factoring a perfect square trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

$\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}$

We can use this equation to factor any perfect square trinomial.

## Perfect square trinomials

A perfect square trinomial can be written as the square of a binomial:

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

Given a perfect square trinomial, factor it into the square of a binomial.

1. Confirm that the first and last term are perfect squares.
2. Confirm that the middle term is twice the product of $\text{\hspace{0.17em}}ab.$
3. Write the factored form as $\text{\hspace{0.17em}}{\left(a+b\right)}^{2}.$

## Factoring a perfect square trinomial

Factor $\text{\hspace{0.17em}}25{x}^{2}+20x+4.$

Notice that $\text{\hspace{0.17em}}25{x}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}4\text{\hspace{0.17em}}$ are perfect squares because $\text{\hspace{0.17em}}25{x}^{2}={\left(5x\right)}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}4={2}^{2}.\text{\hspace{0.17em}}$ Then check to see if the middle term is twice the product of $\text{\hspace{0.17em}}5x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$ The middle term is, indeed, twice the product: $\text{\hspace{0.17em}}2\left(5x\right)\left(2\right)=20x.\text{\hspace{0.17em}}$ Therefore, the trinomial is a perfect square trinomial and can be written as $\text{\hspace{0.17em}}{\left(5x+2\right)}^{2}.$

Factor $\text{\hspace{0.17em}}49{x}^{2}-14x+1.$

${\left(7x-1\right)}^{2}$

## Factoring a difference of squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

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

We can use this equation to factor any differences of squares.

## Differences of squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

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

Given a difference of squares, factor it into binomials.

1. Confirm that the first and last term are perfect squares.
2. Write the factored form as $\text{\hspace{0.17em}}\left(a+b\right)\left(a-b\right).$

## Factoring a difference of squares

Factor $\text{\hspace{0.17em}}9{x}^{2}-25.$

Notice that $\text{\hspace{0.17em}}9{x}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}25\text{\hspace{0.17em}}$ are perfect squares because $\text{\hspace{0.17em}}9{x}^{2}={\left(3x\right)}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}25={5}^{2}.\text{\hspace{0.17em}}$ The polynomial represents a difference of squares and can be rewritten as $\text{\hspace{0.17em}}\left(3x+5\right)\left(3x-5\right).$

Factor $\text{\hspace{0.17em}}81{y}^{2}-100.$

$\left(9y+10\right)\left(9y-10\right)$

Is there a formula to factor the sum of squares?

No. A sum of squares cannot be factored.

## Factoring the sum and difference of cubes

Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.

${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$

Similarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs.

${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$

We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: S ame O pposite A lways P ositive. For example, consider the following example.

${x}^{3}-{2}^{3}=\left(x-2\right)\left({x}^{2}+2x+4\right)$

The sign of the first 2 is the same as the sign between $\text{\hspace{0.17em}}{x}^{3}-{2}^{3}.\text{\hspace{0.17em}}$ The sign of the $\text{\hspace{0.17em}}2x\text{\hspace{0.17em}}$ term is opposite the sign between $\text{\hspace{0.17em}}{x}^{3}-{2}^{3}.\text{\hspace{0.17em}}$ And the sign of the last term, 4, is always positive .

## Sum and difference of cubes

We can factor the sum of two cubes as

${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$

We can factor the difference of two cubes as

${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$

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