Q. \(x^{2}+11x+28\)

Q: Factor the expression x^2+11x+28.

Find two numbers multiplying to 28 and adding to 11: 7 and 4. So x^2+11x+28=(x+7)(x+4).

Q: What are the roots of x^2+11x+28=0?

Using (x+7)(x+4)=0, the roots satisfy x=-7 or x=-4.

Q: Find the vertex of the parabola y=x^2+11x+28.

For y=x^2+11x+28, x-coordinate is -\frac{b}{2a}=-\frac{11}{2}. Then y=\left(-\frac{11}{2}\right)^2+11\left(-\frac{11}{2}\right)+28=\frac{15}{4}.

Q: Does the quadratic have real solutions, and what is the discriminant?

Discriminant D=b^2-4ac=11^2-4(1)(28)=121-112=9, so it has two real solutions.

Answer

We factor the quadratic by finding two numbers that multiply to \(28\) and add to \(11\). Those numbers are \(7\) and \(4\), so the expression factors as

\[
x^2+11x+28=(x+7)(x+4).
\]

Final result: \((x+7)(x+4)\).

Detailed Explanation

We are given the quadratic expression \(x^2 + 11x + 28\). The goal is to simplify or factor it by finding two numbers that multiply to \(28\) and add to \(11\).

Step 1: Identify what we need to factor

Assume the quadratic factors as

\[
x^2 + 11x + 28 = (x + a)(x + b)
\]

When you expand \((x+a)(x+b)\), you get

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

So we need numbers \(a\) and \(b\) such that:

\(a+b = 11\)
\(ab = 28\)

Step 2: Find the numbers \(a\) and \(b\)

We list factor pairs of \(28\):

\(1 \cdot 28 = 28\) and \(1 + 28 = 29\)
\(2 \cdot 14 = 28\) and \(2 + 14 = 16\)
\(4 \cdot 7 = 28\) and \(4 + 7 = 11\)

The pair that works is \(a=4\) and \(b=7\) because \(4+7=11\) and \(4\cdot 7=28\).

Step 3: Write the factored form

Substitute \(a=4\) and \(b=7\) into \((x+a)(x+b)\):

\[
x^2 + 11x + 28 = (x+4)(x+7)
\]

Final Answer

\[
x^2 + 11x + 28 = (x+4)(x+7)
\]

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Algebra FAQ

Factor the expression \(x^2+11x+28\).

Find two numbers multiplying to \(28\) and adding to \(11\): \(7\) and \(4\). So \(x^2+11x+28=(x+7)(x+4)\).

What are the roots of \(x^2+11x+28=0\)?

Using \((x+7)(x+4)=0\), the roots satisfy \(x=-7\) or \(x=-4\).

Complete the square for \(x^2+11x+28\).

\(x^2+11x+28=\left(x+\frac{11}{2}\right)^2-\frac{121}{4}+28=\left(x+\frac{11}{2}\right)^2-\frac{13}{4}\).

Find the vertex of the parabola \(y=x^2+11x+28\).

For \(y=x^2+11x+28\), \(x\)-coordinate is \(-\frac{b}{2a}=-\frac{11}{2}\). Then \(y=\left(-\frac{11}{2}\right)^2+11\left(-\frac{11}{2}\right)+28=\frac{15}{4}\).

Does the quadratic have real solutions, and what is the discriminant?

Discriminant \(D=b^2-4ac=11^2-4(1)(28)=121-112=9\), so it has two real solutions.

Compute the value at \(x=-1\) for the expression \(x^2+11x+28\).

\( (-1)^2+11(-1)+28=1-11+28=18\).

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