Question

factor completely: $x^{2}+13x+40$
$x(x-4)$
$3x(x+6)$
$(x+5)(x+8)$
$5x(x+3)$

Explanation:

Step1: Identify $a, b, c$ for $ax^2+bx+c$

For $x^2+13x+40$, $a=1$, $b=13$, $c=40$

Step2: Find pair with sum $b$, product $ac$

We need two numbers that add to $13$ and multiply to $1\times40=40$. The pair is $5$ and $8$ (since $5+8=13$, $5\times8=40$).

Step3: Rewrite and factor the trinomial

Split the middle term: $x^2+5x+8x+40$
Group and factor: $(x^2+5x)+(8x+40)=x(x+5)+8(x+5)=(x+5)(x+8)$

Answer:

$(x+5)(x+8)$