Chapter 4.6, Problem 6OE

To determine

To Find: Factor of polynomial $y^2+4y-5$

Answer to Problem 6OE

Factor of polynomial $y^2+4y-5=\left(y-1\right)\left(y+5\right)$

Explanation of Solution

Given information: polynomial $y^2+4y-5$

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:

polynomial $y^2+4y-5$

Since the coefficient of $y^2$ is $1$

So, $pr=1$

Therefore, let both pand r equal to $1$

So, that the factorization begins

  $\left(y\right)\left(y\right)$

Since the constant term is $-5$

So, $qs=-5$

Therefore, q and s must have opposite signs. There are two possibilities

  $\left(-1\right)\left(5\right),\left(1\right)\left(-5\right)$

Since the coefficient of y is $4$

So, $ps+qr=4$ , but p and r are both $1$

So, $\begin{array}{l}ps+qr=4\\ s+q=4\end{array}$

Therefore, Choose the pair of factors having sum $4:-1\text{ and 5}$

  $\therefore y^2+4y-5=\left(y-1\right)\left(y+5\right)$