6.2 Factoring a monomial from a polynomial

The product is $3x^7-2x^6+4x^5-3x^4$ and one factor is $x^4$ . Find the other factor.

We have the problem: $x^4$ times "what expression" yields $3x^7-2x^6+4x^5-3x^4$ ? Mathematically,

$x^4\cdot \left(\right)=3x^7-2x^6+4x^5-3x^4$

Since there are four terms in the product, there must be four terms inside the parentheses. To find each of the four terms, we’ll divide (mentally) each term of the product by $x^4$ . The resulting quotient will be the necessary term of the factor.

$\begin{array}{lllll}1stterm:\hfill & \frac{3x^7}{x^4}\hfill & =\hfill & 3x^{7-4}=3x^3\hfill & \text{Place}3x^3\text{into}\text{the}1\text{st}\text{position}\text{in}\text{the}(\begin{array}{cc}& \end{array}).\hfill \\ 2ndterm:\hfill & \frac{-2x^6}{x^4}\hfill & =\hfill & -2x^2\hfill & \text{Place}-2x^2\text{into}\text{the}2\text{nd}\text{position}\text{in}\text{the}(\begin{array}{cc}& \end{array}).\hfill \\ 3rdterm:\hfill & \frac{4x^5}{x^4}\hfill & =\hfill & 4x\hfill & \text{Place}4x\text{into}\text{the}3\text{rd}\text{position}\text{in}\text{the}(\begin{array}{cc}& \end{array}).\hfill \\ 4thterm:\hfill & \frac{-3x^4}{x^4}\hfill & =\hfill & -3\hfill & \text{Place}-3\text{into}\text{the}4\text{th}\text{position}\text{in}\text{the}(\begin{array}{cc}& \end{array}).\hfill \end{array}$

Therefore, the other factor is $3x^3-2x^2+4x-3$ .

This result can be checked by applying the distributive property.

$\begin{array}{rrrr}\hfill x^4\cdot \left(3x^3-2x^2+4x-3\right)& \hfill =& \hfill 3x^7-2x^6+4x^5-3x^4& \hfill \text{Is this correct?}\end{array}$

$\begin{array}{l}\begin{array}{rrrr}\hfill 3x^{4+3}-2x^{4+2}+4x^{4+1}& \hfill =& \hfill 3x^7-2x^6+4x^5-3x^4& \hfill \text{Is this correct?}\end{array}\\ \begin{array}{rrrr}\hfill 3x^7-2x^6+4x^5-3x^4& \hfill =& \hfill 3x^7-2x^6+4x^5-3x^4& \hfill \text{Yes, this is correct}\text{.}\end{array}\end{array}$

Thus,

$x^4\cdot \left(3x^3-2x^2+4x-3\right)=3x^7-2x^6+4x^5-3x^4$

Again, if the divisions can be performed mentally, the process can proceed very quickly.

The product is $10x^3{y}^6+15x^3{y}^4-5x^2{y}^4$ and a factor is $5x^2{y}^4$ . Find the other factor.

$5x^2{y}^4\cdot \left(\right)=10x^3{y}^6+15x^3{y}^4-5x^2{y}^4$

Since there are three terms in the product, there must be three terms inside the parentheses. To find each of these three terms, we’ll divide each term of the product by $5x^2{y}^4$ .

$\begin{array}{lllll}1stterm:\hfill & \frac{10x^3{y}^6}{5x^2{y}^4}\hfill & =\hfill & 2x{y}^2\hfill & \text{Place}\text{the}2x{y}^2\text{into}\text{the}\text{1st}\text{position}\text{in}\text{the}(\begin{array}{cc}& \end{array}).\hfill \\ 2ndterm:\hfill & \frac{15x^3{y}^4}{5x^2{y}^4}\hfill & =\hfill & 3x\hfill & \text{Place}\text{the}3x\text{into}\text{the}\text{2nd}\text{position}\text{in}\text{the}(\begin{array}{cc}& \end{array}).\hfill \\ 3rdterm:\hfill & \frac{-5x^2{y}^4}{5x^2{y}^4}\hfill & =\hfill & -1\hfill & \text{Place}\text{the}-1\text{into}\text{the}\text{3rd}\text{position}\text{in}\text{the}(\begin{array}{cc}& \end{array}).\hfill \end{array}$

The other factor is $2x{y}^2+3x-1$ , and

$5x^2{y}^4\cdot \left(2x{y}^2+3x-1\right)=10x^3{y}^6+15x^3{y}^4-5x^2{y}^4$

The product is $-4a^2-b^3+2c$ and a factor is $-1$ . Find the other factor.

$-1\left(\right)=-4a^2-b^3+2c$

Since there are three terms in the product, there must be three terms inside the parentheses. We will divide (mentally) each term of the product by $-1$ .

$\begin{array}{lllll}1stterm:\hfill & \frac{-4a^2}{-1}\hfill & =\hfill & 4a^2\hfill & \text{Place}4a^2\text{into}\text{the}\text{1st}\text{position}\text{inside}\text{the}(\begin{array}{cc}& \end{array}).\hfill \\ 2ndterm:\hfill & \frac{-b^3}{-1}\hfill & =\hfill & b^3\hfill & \text{Place}{b}^3\text{into}\text{the}\text{2nd}\text{position}\text{inside}\text{the}(\begin{array}{cc}& \end{array}).\hfill \\ 3rdterm:\hfill & \frac{2c}{-1}\hfill & =\hfill & -2c\hfill & \text{Place}-2c\text{into}\text{the}\text{3rd}\text{position}\text{inside}\text{the}(\begin{array}{cc}& \end{array}).\hfill \end{array}$

The other factor is $4a^2+b^3-2c$ , and

$-1(4a^2+b^3-2c)=-4a^2-b^3+2c$

Without writing the $-1$ , we get

$-(4a^2+b^3-2c)=-4a^2-b^3+2c$

The product is $-3a^2{b}^5-15a^3{b}^2+9a^2{b}^2$ and a factor is $-3a^2{b}^2$ . Find the other factor.

$-3a^2{b}^2\left(\right)=-3a^2{b}^5-15a^3{b}^2+9a^2{b}^2$

Mentally dividing each term of the original trinomial by $-3a^2{b}^2$ , we get $b^3+5a-3$ as the other factor, and

$-3a^2{b}^2(b^3+5a-3)=-3a^2{b}^5-15a^3{b}^2+9a^2{b}^2$