Question

assignment 5.4 dividing polynomials
due friday by 11:59pm points 9 submitting an external tool
assignment 5.4 dividing polynomials
score: 1/9 answered: 1/9
question 2
find the quotient and remainder using polynomial long division.
\\(\frac{x^{3}-7x^{2}+40}{x - 6}\\)
the quotient is
the remainder is
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Explanation:

Step1: Divide leading terms

Divide $x^{3}$ by $x$. We get $x^{2}$. Multiply $x - 6$ by $x^{2}$: $x^{2}(x - 6)=x^{3}-6x^{2}$. Subtract from the dividend: $(x^{3}-7x^{2}+40)-(x^{3}-6x^{2})=-x^{2}+40$.

Step2: Divide new leading term

Divide $-x^{2}$ by $x$. We get $-x$. Multiply $x - 6$ by $-x$: $-x(x - 6)=-x^{2}+6x$. Subtract from the new dividend: $(-x^{2}+40)-(-x^{2}+6x)=-6x + 40$.

Step3: Divide new leading term again

Divide $-6x$ by $x$. We get $-6$. Multiply $x - 6$ by $-6$: $-6(x - 6)=-6x+36$. Subtract from the new dividend: $(-6x + 40)-(-6x+36)=4$.

Answer:

The quotient is $x^{2}-x - 6$
The remainder is $4$