$$ \left(3 x^{2}+2 x-3\right) \div(x+3)=\square+(\square) $$ (Type the quotient in the first answer box and the

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UpStudy AI · Accepted Answer

We want to divide
$$
3x^2 + 2x - 3
$$
by
$$
x+3.
$$

**Step 1. Divide the leading term.**

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

**Step 2. Multiply and subtract.**

Multiply $$3x$$ by the divisor $$x+3$$:
$$
3x \cdot (x+3) = 3x^2 + 9x.
$$
Subtract this from the original polynomial:
$$
(3x^2 + 2x - 3) - (3x^2 + 9x) = (3x^2-3x^2) + (2x-9x) - 3 = -7x - 3.
$$

**Step 3. Divide the new leading term.**

Divide $$-7x$$ by $$x$$:
$$
\frac{-7x}{x} = -7.
$$
This is the next term of the quotient.

**Step 4. Multiply and subtract again.**

Multiply $$-7$$ by the divisor:
$$
-7 \cdot (x+3) = -7x - 21.
$$
Subtract this product from $$-7x - 3$$:
$$
(-7x - 3) - (-7x - 21) = -7x - 3 + 7x + 21 = 18.
$$
The number $$18$$ is the remainder.

**Final Answer**

Thus,
$$
\frac{3x^2+2x-3}{x+3}=3x-7+\frac{18}{x+3}.
$$

So the quotient is $$3x-7$$ and the remainder is $$\frac{18}{x+3}$$.
The quotient is $$3x - 7$$ and the remainder is $$\frac{18}{x+3}$$.

(Type the quotient in the first answer box and the

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