Answer:
[tex]\boxed{\boxed{\sf{(x-3)(x+16)~or~(x+16)(x-3)}}}[/tex]
Solution Steps:
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1.) Factor the expression by grouping:
- First, the expression needs to be rewritten as [tex]x^2+ax+bx-48[/tex]. To find a and b, set up a system to be solved:
2.) List all such integer pairs that give product −48:
3.) Calculate the sum for each pair:
4.) The solution is the pair that gives sum 13:
5.) Rewrite [tex]\bold{x^2+13x-48}[/tex]:
- [tex]x^2+13x-48=(x^2-3x)+(16x-48)[/tex]
6.) Factor out [tex]\bold{(x)}[/tex] in the first and 16 in the second group:
- [tex]x(x-3)+16(x-3)[/tex]
7.) Factor out common term [tex]\bold{x-3}[/tex] by using distributive property:
So your answer is [tex]\bold{(x-3)(x+16)}[/tex] or [tex]\bold{(x+16)(x-3)}[/tex].
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