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How do you factor #5x^3+6x^2-45x-54#?

Answer 1

Ethan Adkins

(5x +6)(x-3)(x+3)

Group the terms in 'pairs'

thus #[5x^3+6x^2]+[-45x-54]#

now factorise each pair

#rArrx^2(5x+6)-9(5x+6)#

We now have a common factor of (5x + 6) and taking it out leaves.

#(5x+6)(x^2-9)#

Now #x^2-9" is a difference of squares"#

In general a difference of squares factorises as.

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

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

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

Putting this altogether to obtain.

#5x^3+6x^2-45x-54=(5x+6)(x-3)(x+3)#

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Answer 2

Ethan Adkins

To factor the expression 5x^3 + 6x^2 - 45x - 54, you can first look for any common factors among the terms, then use synthetic division or long division to find the roots, and finally apply the factor theorem to factorize the polynomial completely.

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Answer from HIX Tutor

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