identify the quotient and the remainder.\n$(24x^{3}-14x^{2}+20x + 6)div(4x^{2}-3x + 5)=q+\frac{r}{4x^{2}-3x…
identify the quotient and the remainder.\n$(24x^{3}-14x^{2}+20x + 6)div(4x^{2}-3x + 5)=q+\frac{r}{4x^{2}-3x + 5}\nq=\nr=$
Answer
Answer:
$Q = 6x + 1$ $R=-7x + 1$
Explanation:
Step1: Divide leading terms
Divide $24x^{3}$ by $4x^{2}$, we get $6x$.
Step2: Multiply divisor by $6x$
$(4x^{2}-3x + 5)\times6x=24x^{3}-18x^{2}+30x$.
Step3: Subtract
$(24x^{3}-14x^{2}+20x + 6)-(24x^{3}-18x^{2}+30x)=4x^{2}-10x + 6$.
Step4: Divide new leading term
Divide $4x^{2}$ by $4x^{2}$, we get $1$.
Step5: Multiply divisor by $1$
$(4x^{2}-3x + 5)\times1=4x^{2}-3x + 5$.
Step6: Subtract again
$(4x^{2}-10x + 6)-(4x^{2}-3x + 5)=-7x + 1$. So the quotient $Q = 6x+1$ and the remainder $R=-7x + 1$.