Question

Rewrite each expression as a polynomial in standard form: x^2-5x+6 ÷ x-3 + x^3-1 ÷ x-1

Answer 1

Answer:

So in standard form equation will be $x^2+2x-1$

Step-by-step explanation:

We have given expression $\frac{x^2-5x+6}{x-3}+\frac{x^3-1}{x-1}$

Let first we solve first part of the expression

So $\frac{x^2-5x+6}{x-3}=\frac{x^2-3x-2x+6}{x-3}=\frac{(x-3)(x-2)}{x-3}=x-2$

Now second part $\frac{x^3-1}{x-1}$

We know the algebraic identity $a^3-b^3=(a-b)(a^2+ab+b^2)$

So by using this identity

$\frac{x^3-1}{x-1}=\frac{(x-1)(x^2+x+1)}{x-1}=x^2+x+1$

Now adding first and second part

$x-2+x^2+x+1=x^2+2x-1$

So in standard form equation will be $x^2+2x-1$