Chapter 4.6, Problem 15OE

To determine

To Find: The other factor of $x^2+3x-10$

Answer to Problem 15OE

The other factor of $x^2+3x-10$ is $\left(x+5\right)$

Explanation of Solution

Given information: One factor of $x^2+3x-10$ is $\left(x-2\right)$

Formula used: If the quadratic trinomial $a{x}^2+bx+c$ can be factored into the product $\left(px+q\right)\left(rx+s\right)$ where $p,q,r,s$ are integers, then

  $a{x}^2+bx+c=pr{x}^2+\left(ps+qr\right)x+qs$

Setting corresponding coefficients equal gives

  $a=pr,b=ps+qr,c=qs$

Calculation:

One factor of $x^2+3x-10$ is $\left(x-2\right)$

Let the other factor is $\left(rx+s\right)$

Then $x^2+3x-10=\left(x-2\right)\left(rx+s\right)$

where $r,s$ are integers, then

  $x^2+3x-10=r{x}^2+\left(s-2r\right)x-2s$

Setting corresponding coefficients equal gives

  $r=1,s=5$

The other factor of $x^2+3x-10$ is $\left(rx+s\right)$

Put $r=1,s=5$

The other factor of $x^2+3x-10$ is $\left(x+5\right)$