Question

Factor the trinomial

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

Short answer

The solution$2{\left(a+b\right)}^2+5\left(a+b\right)-3=\left(2\left(a+b\right)-1\right)\left(\left(a+b\right)+3\right)$

Step-by-step solution

Step 1. Apply the concept of factoring trinomials.

To factor a trinomial of the form $a{x}^2+bx+c;a\ne 0$, we note that

$a{x}^2+bx+c=\left(px+r\right)\left(qx+s\right)=pq{x}^2+\left(ps+qr\right)x+rs$

So, we need to choose numbers p, q, r and s so that $pq=a$, $ps+qr=b$and $rs=c$.

Step 2. Analyze the trinomial.

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

Let $a+b=y$

So, the given trinomial becomes:

$2y^2+5y-3$

This is of the form $a{y}^2+by+c$where $a=2,b=5,c=-3$

The factorization of the trinomial will be $\left(py+r\right)\left(qy+s\right)=pq{y}^2+\left(ps+qr\right)y+rs$

So, we need to choose numbers p, q, r and s so that $pq=a$, $ps+qr=b$and $rs=c$.

Step 3. Find the integers to factor the trinomial.

We need to choose integers r and s so that $pq=2,ps+qr=5,rs=-3$

By trial and error, we find that the two integers are $p=2,q=1,r=-1,s=3$

$pq=2$

role="math" $ps+qr=6-1=5$

$rs=(-1)\cdot 3=-3$

Step 4. Factor the trinomial.

The factorization of the trinomial is

$\begin{array}{l}2y^2+5y-3=\left(py+r\right)\left(qy+s\right)\\ \Rightarrow 2y^2+5y-3=\left(2y-1\right)\left(y+3\right)\\ \Rightarrow 2{\left(a+b\right)}^2+5\left(a+b\right)-3=\left(2\left(a+b\right)-1\right)\left(\left(a+b\right)+3\right)\mathrm{..........}\left[y=\left(a+b\right)\right]\\ \Rightarrow 2{\left(a+b\right)}^2+5\left(a+b\right)-3=\left(2\left(a+b\right)-1\right)\left(\left(a+b\right)+3\right)\end{array}$