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How do you factor #x^3 + x^2 -9x -9# by grouping?

Answer 1

Hazel Anglin

(x+1)(x-3)(x+3)

Group the terms into 'pairs' as follows.

#[x^3+x^2]+[-9x-9]#

now factorise each pair.

#color(red)(x^2)(x+1)-color(red)(9)(x+1)#

We now have a common factor of (x+1) which can be 'taken out'

#(x+1)(color(red)(x^2-9))........ (A)#

#color(red)(x^2-9)" is a " color(blue)"difference of squares"# and is factorised in general as follows.

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))#

#x^2=(x)^2" and " 9=(3)^2rArra=x" and " b=3#

#rArrx^2-9=(x-3)(x+3)#

Substitute these factors into (A)

#rArr(x+1)(x^2-9)=(x+1)(x-3)(x+3)#

Thus #x^3+x^2-9x-9=(x+1)(x-3)(x+3)#

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Answer 2

Ethan Adkins

To factor the expression x^3 + x^2 - 9x - 9 by grouping, you can first group the terms in pairs:

(x^3 + x^2) + (-9x - 9)

Then, factor out the common terms from each pair:

x^2(x + 1) - 9(x + 1)

Now, you can see that both terms share a common factor of (x + 1), so you can factor it out:

(x + 1)(x^2 - 9)

Further factor the quadratic term using the difference of squares:

(x + 1)(x - 3)(x + 3)

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Answer from HIX Tutor

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