Question

Jay factored the 4 term polynomial : x^3 - 9x +2x^2 - 18 and decided that the complete factorization was : ( x + 2 ) (x^2-9 ). Before turning in his paper, he checked his final factorization by multiplying out his factors and was sure that he had found the correct factors. When his teacher graded his paper, she marked his answer as incorrect, but gave him one more chance to show the correct factorization.
a) What was Jay's mistake?
b) Show/Describe EACH step in factoring the 4 term expression correctly and completely : x^3 - 9x +2x^2 - 18

Answer 1

Answer:

$x^3-9x+2x^2-18\left(x+2\right)\left(x+3\right)\left(x-3\right)$

Step-by-step explanation:

we are given that $x^3-9x+2x^2-18$

w are sked to step by step factorise the above polynomial

$\left(x^3+2x^2\right)+\left(-9x-18\right)$

$-9\mathrm{\:from\:}-9x-18\mathrm{:\quad }-9\left(x+2\right)$

$-9x-9\cdot \:2$

$-9\left(x+2\right)$

$\mathrm{Factor\:out\:}x^2\mathrm{\:from\:}x^3+2x^2\mathrm{:\quad }x^2\left(x+2\right)$

$-9\left(x+2\right)+x^2\left(x+2\right)$

$\left(x+2\right)\left(x^2-9\right)$

$x^2-9:\quad \left(x+3\right)\left(x-3\right)$

$x^2-9$

$\mathrm{Rewrite\:}9\mathrm{\:as\:}3^2$

$=x^2-3^2$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$x^2-3^2=\left(x+3\right)\left(x-3\right)$

Hence

$x^3-9x+2x^2-18=\left(x+2\right)\left(x+3\right)\left(x-3\right)$

a) The jay mistake was he did not factorise  $x^3-9x+2x^2-18x^2-9 furtherb) the complete answer wil be [tex]x^3-9x+2x^2-18\left(x+2\right)\left(x+3\right)\left(x-3\right)$

Answer 2

Answer:

$x^3 + 2x^2 - 9x - 18 = (x+3)(x-3)(x+2)$  

Step-by-step explanation:

a) Jay's factorization was correct.

$x^3 + 2x^2 - 9x - 18\\=(x^2-9)(x+2)$

Jay' mistake was that he further did not factorized the factor $(x^2-9)$, which can be further broken into factors.

$x^3 + 2x^2 - 9x - 18\\=x^2(x+2) - 9(x+2)\\=(x^2-9)(x+2)\\=(x+3)(x-3)(x+2)$

We used the formula:

$(x^2 - y^2) = (x+y)(x-y)$ to factorize the term $(x^2 - 9)$