Respuesta :
Answer:
[tex]x^3-9x+2x^2-18\left(x+2\right)\left(x+3\right)\left(x-3\right)[/tex]
Step-by-step explanation:
we are given that [tex]x^3-9x+2x^2-18[/tex]
w are sked to step by step factorise the above polynomial
[tex]\left(x^3+2x^2\right)+\left(-9x-18\right)[/tex]
[tex]-9\mathrm{\:from\:}-9x-18\mathrm{:\quad }-9\left(x+2\right)[/tex]
[tex]-9x-9\cdot \:2[/tex]
[tex]-9\left(x+2\right)[/tex]
[tex]\mathrm{Factor\:out\:}x^2\mathrm{\:from\:}x^3+2x^2\mathrm{:\quad }x^2\left(x+2\right)[/tex]
[tex]-9\left(x+2\right)+x^2\left(x+2\right)[/tex]
[tex]\left(x+2\right)\left(x^2-9\right)[/tex]
[tex]x^2-9:\quad \left(x+3\right)\left(x-3\right)[/tex]
[tex]x^2-9[/tex]
[tex]\mathrm{Rewrite\:}9\mathrm{\:as\:}3^2[/tex]
[tex]=x^2-3^2[/tex]
[tex]\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)[/tex]
[tex]x^2-3^2=\left(x+3\right)\left(x-3\right)[/tex]
Hence
[tex]x^3-9x+2x^2-18=\left(x+2\right)\left(x+3\right)\left(x-3\right)[/tex]
a) The jay mistake was he did not factorise [tex]x^3-9x+2x^2-18x^2-9 further
b) the complete answer wil be
[tex]x^3-9x+2x^2-18\left(x+2\right)\left(x+3\right)\left(x-3\right)[/tex]
Answer:
[tex]x^3 + 2x^2 - 9x - 18 = (x+3)(x-3)(x+2)[/tex]
Step-by-step explanation:
a) Jay's factorization was correct.
[tex]x^3 + 2x^2 - 9x - 18\\=(x^2-9)(x+2)[/tex]
Jay' mistake was that he further did not factorized the factor [tex](x^2-9)[/tex], which can be further broken into factors.
[tex]x^3 + 2x^2 - 9x - 18\\=x^2(x+2) - 9(x+2)\\=(x^2-9)(x+2)\\=(x+3)(x-3)(x+2)[/tex]
We used the formula:
[tex](x^2 - y^2) = (x+y)(x-y)[/tex] to factorize the term [tex](x^2 - 9)[/tex]
