Q. What is the quotient of \(x^2 + 7x + 12\) and \(x + 4\)?

Q: What is the quotient of x^2 + 7x + 12 and x + 4?

The quotient is x + 3 with remainder 0, because x^2+7x+12=(x+4)(x+3). Note the original expression is undefined at x=-4.

Q: How do you get that by factoring?

Factor: x^2+7x+12=(x+3)(x+4). Cancel the common factor x+4 to get quotient x+3, provided x\neq-4.

Q: How do you get that by synthetic division?

Use root -4. Coefficients 1, 7, 12 give synthetic steps: bring 1, multiply -4 → -4, add → 3, multiply -4 → -12, add → 0. Quotient coefficients 1 and 3 → x+3.

Q: Is there a remainder and what does remainder 0 mean?

Remainder 0 means the divisor x+4 divides the polynomial exactly, so x+4 is a factor and the division yields an exact polynomial quotient x+3.

Q: Why must we state x\ne-4 after cancellation?

Cancelling x+4 removes the factor algebraically, but division by zero is undefined. The simplified form x+3 is valid for all x\neq-4; the original expression is not defined at x=-4.

Answer

\[
x^2+7x+12=(x+3)(x+4)
\]

\[
\frac{x^2+7x+12}{x+4}=x+3.
\]

Detailed Explanation

Write the division to be performed: divide
\(x^{2} + 7x + 12\) by \(x + 4\).
Divide the leading term of the dividend by the leading term of the divisor.
Compute \(\dfrac{x^{2}}{x} = x\).
This gives the first term of the quotient: \(x\).
Multiply the divisor by that quotient term:
\(x\cdot (x + 4) = x^{2} + 4x\).
Subtract this product from the dividend:
\((x^{2} + 7x + 12) – (x^{2} + 4x) = 3x + 12\).
The \(x^{2}\) terms cancel, leaving \(3x + 12\).
Divide the leading term of the new remainder by the leading term of the divisor:
\(\dfrac{3x}{x} = 3\).
This gives the next term of the quotient: \(3\).
Multiply the divisor by 3: \(3\cdot (x + 4) = 3x + 12\).
Subtract this product from the current remainder:
\((3x + 12) – (3x + 12) = 0\).
The remainder is 0, so the division is exact.
Conclude that the quotient is
\(x + 3\).
(Alternative check by factoring) Factor the quadratic:
\(x^{2} + 7x + 12 = (x + 3)(x + 4)\).
Dividing by \(x + 4\) leaves \(x + 3\), confirming the quotient.

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FAQs

What is the quotient of \(x^2 + 7x + 12\) and \(x + 4\)?

The quotient is \(x + 3\) with remainder \(0\), because \(x^2+7x+12=(x+4)(x+3)\). Note the original expression is undefined at \(x=-4\).

How do you get that by factoring?

Factor: \(x^2+7x+12=(x+3)(x+4)\). Cancel the common factor \(x+4\) to get quotient \(x+3\), provided \(x\neq-4\).

How do you get that by synthetic division?

Use root \(-4\). Coefficients 1, 7, 12 give synthetic steps: bring 1, multiply -4 → -4, add → 3, multiply -4 → -12, add → 0. Quotient coefficients 1 and 3 → \(x+3\).

Is there a remainder and what does remainder 0 mean?

Remainder \(0\) means the divisor \(x+4\) divides the polynomial exactly, so \(x+4\) is a factor and the division yields an exact polynomial quotient \(x+3\).

Why must we state \(x\ne-4\) after cancellation?

Cancelling \(x+4\) removes the factor algebraically, but division by zero is undefined. The simplified form \(x+3\) is valid for all \(x\neq-4\); the original expression is not defined at \(x=-4\).

How can I check the result quickly?

What if the divisor were not a factor (e.g., \(x+5\))?

Then division gives a nonzero remainder. Use polynomial long division or synthetic division to find quotient and remainder; express as quotient + remainder/(divisor).

Divide x^2+7x+12 by x+4 to get x+3.
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