Question

Please help!!!!!!!:>

Answer 1

Answer:

answer is (x+16)(x-3) I'm sure

Answer 2

Answer:

$\boxed{\boxed{\sf{(x-3)(x+16)~or~(x+16)(x-3)}}}$

Solution Steps:

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1.) Factor the expression by grouping:

 - First, the expression needs to be rewritten as $x^2+ax+bx-48$. To find a and b, set up a system to be solved:

• $a+b=13$

• $ab=1(-48)=-48$

2.) List all such integer pairs that give product −48:

• $-1,48$

• $-2,24$

• $-3,16$

• $-4,12$

• $-6,8$

3.) Calculate the sum for each pair:

• $-1+48=47$

• $-2+24=22$

• $-3+16=13$

• $-4+12=8$

• $-6+8=2$

4.) The solution is the pair that gives sum 13:

• $a=-3$

• $b=16$

5.) Rewrite $\bold{x^2+13x-48}$:

• $x^2+13x-48=(x^2-3x)+(16x-48)$

6.) Factor out $\bold{(x)}$ in the first and 16 in the second group:

• $x(x-3)+16(x-3)$

7.) Factor out common term $\bold{x-3}$ by using distributive property:

• $(x-3)(x+16)$

So your answer is $\bold{(x-3)(x+16)}$ or $\bold{(x+16)(x-3)}$.

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