1 Answer

Jitendra Gaur · Answer 1 · 2021-02-03T10:37:32+0000

Hello,

First of all, dividing by 0 is not defined we will take x different from 1/3 (because -1+3x= 0 <=> x = 1/3)

Then, dividing this polynomial by (-1+3x) means that you have to have Q(x), a polynomial of degree 1 and R a real number such that

$(9x^2-6x+2)/(-1+3x)=Q(x)+(R)/(-1+3x)\\\\\text{We can multiply by (-1+3x)}\\\\9x^2-6x+2=(-1+3x)Q(x)+R$

Either you do the division and you can select the correct answer or you check the different answer to identify the correct one.

For instance, for the first answer, let's estimate

$(-1+3x)(3x-3)+(-1+3x)/(3x-1)\\\\=-3x+3+9x^2-9x+1\\\\=9x^2-12x+4$

and this is not equal to our initial polynomial, so this is not the correct answer.

Second Answer.

$(-1+3x)(3x-1)+(-1+3x)/(3x-1)\\\\=-3x+1+9x^2-3x+1\\\\=9x^2-6x+2$

And this is our initial polynomial. So this is the
$\large \boxed{\sf \bf \text{correct answer}}$

Thank you.

PS: in this example, you can even notice the following (so it makes the division trivial)

$(3x-1)^x=9x^2-6x+1 \\\\\text{So}\\\\9x^2-6x+2= (3x-1)^2+1\\\\\text{And, then.}\\\\(9x^2-6x+2)/(-1+3x)=3x-1+(1)/(3x-1)$

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