Divide. $$ \left(12 x^{3}-19 x^{2}+16 x+5\right) \div\left(4 x^{2}-x\right) $$ Your answer should give the quotient and the remainder: Quotient: Remainder:

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UpStudy AI · Accepted Answer

**Step 1:** Identify the dividend and the divisor.

We have the dividend
$$
12x^3 - 19x^2 + 16x + 5
$$
and the divisor
$$
4x^2 - x.
$$

**Step 2:** Divide the leading term of the dividend by the leading term of the divisor.

Divide $$12x^3$$ (from the dividend) by $$4x^2$$ (from the divisor):
$$
\frac{12x^3}{4x^2} = 3x.
$$
This gives the first term of the quotient.

**Step 3:** Multiply the entire divisor by $$3x$$ and subtract from the dividend.

Multiply the divisor by $$3x$$:
$$
3x \cdot \left(4x^2 - x\right) = 12x^3 - 3x^2.
$$
Subtract this from the dividend:
$$
\begin{array}{rcl}
(12x^3 - 19x^2 + 16x + 5) - (12x^3 - 3x^2) &=& -16x^2 + 16x + 5.
\end{array}
$$

**Step 4:** Divide the new leading term $$-16x^2$$ by $$4x^2$$.

Calculate:
$$
\frac{-16x^2}{4x^2} = -4.
$$
This gives the next term of the quotient.

**Step 5:** Multiply the entire divisor by $$-4$$ and subtract.

Multiply:
$$
-4 \cdot \left(4x^2 - x\right) = -16x^2 + 4x.
$$
Subtract this product from the previous remainder:
$$
\begin{array}{rcl}
(-16x^2 + 16x + 5) - (-16x^2 + 4x) &=& (16x - 4x) + 5 \
&=& 12x + 5.
\end{array}
$$
Since the degree of $$12x + 5$$ is less than the degree of the divisor $$4x^2 - x$$, this is the final remainder.

**Final Answer:**

Quotient: $$3x - 4$$  
Remainder: $$12x + 5$$
Quotient: $$3x - 4$$ Remainder: $$12x + 5$$

Divide.

Your answer should give the quotient and the remainder:
Quotient:
Remainder:

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