Question

assignment 5.4 dividing polynomials
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assignment 5.4 dividing polynomials
score: 0/9 answered: 0/9
question 1
find the quotient and remainder using polynomial long division.
\\(\frac{18x^{2}-6x - 23}{3x - 4}\\)
the quotient is
the remainder is
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Explanation:

Step1: Divide leading - terms

Divide the leading term of the numerator $18x^{2}$ by the leading term of the denominator $3x$. $\frac{18x^{2}}{3x}=6x$. This is the first term of the quotient.

Step2: Multiply and subtract

Multiply the denominator $3x - 4$ by $6x$: $6x(3x - 4)=18x^{2}-24x$. Subtract this from the numerator: $(18x^{2}-6x - 23)-(18x^{2}-24x)=18x^{2}-6x - 23 - 18x^{2}+24x = 18x-23$.

Step3: Divide leading - terms again

Divide the leading term of the new numerator $18x$ by the leading term of the denominator $3x$. $\frac{18x}{3x}=6$. This is the second term of the quotient.

Step4: Multiply and subtract again

Multiply the denominator $3x - 4$ by $6$: $6(3x - 4)=18x-24$. Subtract this from the new numerator: $(18x-23)-(18x - 24)=18x-23 - 18x + 24=1$.

Answer:

The quotient is $6x + 6$, the remainder is $1$