Q. \(x^2 + 5x – 24\)

Q: How do you factor x^2+5x-24?

Find two numbers with product -24 and sum 5: 8 and -3. So x^2+5x-24=(x+8)(x-3).

Q: What are the roots of x^2+5x-24=0?

Set (x+8)(x-3)=0. Then x=-8 or x=3.

Q: What is the discriminant of x^2+5x-24?

For ax^2+bx+c, \Delta=b^2-4ac. Here \Delta=5^2-4(1)(-24)=25+96=121.

Q: What is x^2+5x-24 in completed square form?

Write x^2+5x=(x+\frac{5}{2})^2-\frac{25}{4}. Then x^2+5x-24=(x+\frac{5}{2})^2-\frac{121}{4}.

Q: What is the vertex of y=x^2+5x-24?

For y=ax^2+bx+c, vertex x-value is -\frac{b}{2a}=-\frac{5}{2}. Then y=(-\frac{5}{2})^2+5(-\frac{5}{2})-24=\frac{25}{4}-\frac{25}{2}-24=-\frac{121}{4}.

Answer

Factor the quadratic by finding two numbers that multiply to \(-24\) and add to \(5\). Those numbers are \(8\) and \(-3\).

\[
x^2+5x-24=(x+8)(x-3)
\]

Set each factor equal to zero:

\[
x+8=0 \Rightarrow x=-8,\quad x-3=0 \Rightarrow x=3
\]

Final result: \((x+8)(x-3)=0\), so \(x=-8\) or \(x=3\).

Detailed Explanation

We want to simplify the expression \(x^2 + 5x – 24\). A common first goal is to factor it, since factoring helps find useful structure (like roots).

Step 1: Set up factoring

We look for two numbers \(m\) and \(n\) such that

\[
x^2 + 5x – 24 = (x + m)(x + n).
\]

When you expand \((x+m)(x+n)\), you get

\[
(x+m)(x+n) = x^2 + (m+n)x + mn.
\]

So we need:

\(m+n = 5\)
\(mn = -24\)

Step 2: Find the numbers

We need two numbers whose product is \(-24\) and whose sum is \(5\).

List factor pairs of \(-24\):

\(1\) and \(-24\) (sum \(=-23\))
\(2\) and \(-12\) (sum \(=-10\))
\(3\) and \(-8\) (sum \(=-5\))
\(4\) and \(-6\) (sum \(=-2\))
\(-1\) and \(24\) (sum \(=23\))
\(-2\) and \(12\) (sum \(=10\))
\(-3\) and \(8\) (sum \(=5\))

The pair \(-3\) and \(8\) works because

\((-3) + 8 = 5\)
\((-3)\cdot 8 = -24\)

Step 3: Write the factored form

Substitute \(m=-3\) and \(n=8\) into \((x+m)(x+n)\):

\[
x^2 + 5x – 24 = (x – 3)(x + 8).
\]

Final Answer

\[
x^2 + 5x – 24 = (x – 3)(x + 8).
\]

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Algebra FAQ

How do you factor \(x^2+5x-24\)?

Find two numbers with product \(-24\) and sum \(5\): \(8\) and \(-3\). So \(x^2+5x-24=(x+8)(x-3)\).

What are the roots of \(x^2+5x-24=0\)?

Set \((x+8)(x-3)=0\). Then \(x=-8\) or \(x=3\).

Can you use the quadratic formula to solve \(x^2+5x-24=0\)?

\(x=\frac{-5\pm\sqrt{5^2-4(1)(-24)}}{2}=\frac{-5\pm\sqrt{25+96}}{2}=\frac{-5\pm 11}{2}\). So \(x=3\) or \(x=-8\).

What is the discriminant of \(x^2+5x-24\)?

For \(ax^2+bx+c\), \(\Delta=b^2-4ac\). Here \(\Delta=5^2-4(1)(-24)=25+96=121\).

What is \(x^2+5x-24\) in completed square form?

Write \(x^2+5x=(x+\frac{5}{2})^2-\frac{25}{4}\). Then \(x^2+5x-24=(x+\frac{5}{2})^2-\frac{121}{4}\).

What is the vertex of \(y=x^2+5x-24\)?

For \(y=ax^2+bx+c\), vertex \(x\)-value is \(-\frac{b}{2a}=-\frac{5}{2}\). Then \(y=(-\frac{5}{2})^2+5(-\frac{5}{2})-24=\frac{25}{4}-\frac{25}{2}-24=-\frac{121}{4}\).

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