ashhhhh114
Answered

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Please help!!!!!!!:>

Please Helpgt class=

Sagot :

frankscm310

Answer:

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

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:

  • [tex]a+b=13[/tex]
  • [tex]ab=1(-48)=-48[/tex]

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

  • [tex]-1,48[/tex]
  • [tex]-2,24[/tex]
  • [tex]-3,16[/tex]
  • [tex]-4,12[/tex]
  • [tex]-6,8[/tex]

3.) Calculate the sum for each pair:

  • [tex]-1+48=47[/tex]
  • [tex]-2+24=22[/tex]
  • [tex]-3+16=13[/tex]
  • [tex]-4+12=8[/tex]
  • [tex]-6+8=2[/tex]

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

  • [tex]a=-3[/tex]
  • [tex]b=16[/tex]

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:

  • [tex](x-3)(x+16)[/tex]

So your answer is [tex]\bold{(x-3)(x+16)}[/tex] or [tex]\bold{(x+16)(x-3)}[/tex].

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