Q. \(x^2 + 11x + 24\)

Q: Factor x^2+11x+24.

Use two numbers with sum 11 and product 24: 3 and 8. So x^2+11x+24=(x+3)(x+8).

Q: Solve x^2+11x+24=0.

From (x+3)(x+8)=0, set each factor to zero: x=-3 or x=-8.

Q: Find the x-intercepts of y=x^2+11x+24.

Set y=0: x=-3 and x=-8. Intercepts are (-3,0) and (-8,0).

Q: What is the vertex of y=x^2+11x+24?

For ax^2+bx+c, x=-\frac{b}{2a}=-\frac{11}{2}. Then y=\left(-\frac{11}{2}\right)^2+11\left(-\frac{11}{2}\right)+24=-\frac{25}{4}.

Q: Complete the square for x^2+11x+24.

Group: x^2+11x=(x+\frac{11}{2})^2-\frac{121}{4}. Then add 24=\frac{96}{4}: x^2+11x+24=(x+\frac{11}{2})^2-\frac{25}{4}.

Q: Determine if the quadratic opens up or down, and the minimum value.

Since a=1>0, it opens up. The minimum value is the vertex y=-\frac{25}{4}.

Answer

We factor the quadratic \(x^2+11x+24\). Find two numbers that multiply to \(24\) and add to \(11\): \(3\) and \(8\).

\[
x^2+11x+24=(x+3)(x+8)
\]

So the factored form is \((x+3)(x+8)\), and the zeros are \(x=-3\) and \(x=-8\).

Detailed Explanation

We want to simplify or factor the quadratic expression \(x^2+11x+24\). A common approach is to factor it into the form \((x+a)(x+b)\).

Step 1: Identify what we need to match.

Suppose

\[
x^2+11x+24=(x+a)(x+b).
\]

Expanding the right-hand side gives

\[
(x+a)(x+b)=x^2+(a+b)x+ab.
\]

So we need to match coefficients:

The coefficient of \(x\) must be \(a+b=11\).
The constant term must be \(ab=24\).

Step 2: Find two numbers with product \(24\) and sum \(11\).

We list factor pairs of \(24\):

\(1 \cdot 24 = 24\) and \(1+24=25\)
\(2 \cdot 12 = 24\) and \(2+12=14\)
\(3 \cdot 8 = 24\) and \(3+8=11\)

The pair \(3\) and \(8\) works because \(3+8=11\) and \(3\cdot 8=24\).

Step 3: Write the factored form.

Substitute \(a=3\) and \(b=8\):

\[
x^2+11x+24=(x+3)(x+8).
\]

Final Answer:

\[
x^2+11x+24=(x+3)(x+8).
\]

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Algebra FAQ

Factor \(x^2+11x+24\).

Use two numbers with sum \(11\) and product \(24\): \(3\) and \(8\). So \(x^2+11x+24=(x+3)(x+8)\).

Solve \(x^2+11x+24=0\).

From \((x+3)(x+8)=0\), set each factor to zero: \(x=-3\) or \(x=-8\).

Find the \(x\)-intercepts of \(y=x^2+11x+24\).

Set \(y=0\): \(x=-3\) and \(x=-8\). Intercepts are \((-3,0)\) and \((-8,0)\).

What is the vertex of \(y=x^2+11x+24\)?

For \(ax^2+bx+c\), \(x=-\frac{b}{2a}=-\frac{11}{2}\). Then \(y=\left(-\frac{11}{2}\right)^2+11\left(-\frac{11}{2}\right)+24=-\frac{25}{4}\).

Complete the square for \(x^2+11x+24\).

Group: \(x^2+11x=(x+\frac{11}{2})^2-\frac{121}{4}\). Then add \(24=\frac{96}{4}\): \(x^2+11x+24=(x+\frac{11}{2})^2-\frac{25}{4}\).

Determine if the quadratic opens up or down, and the minimum value.

Since \(a=1>0\), it opens up. The minimum value is the vertex \(y=-\frac{25}{4}\).

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