Q. \(x^2+4x-12\)

Q: What are the solutions to x^2+4x-12=0?

Factor x^2+4x-12=(x+6)(x-2). So x=-6 or x=2.

Q: Can you complete the square for x^2+4x-12?

x^2+4x-12=(x+2)^2-16. Then (x+2)^2=16, giving x=2 or x=-6.

Q: What is the vertex of y=x^2+4x-12?

For y=x^2+4x-12, x-coordinate is -\frac{b}{2a}=-\frac{4}{2}=-2. Vertex is at (-2,( -2)^2+4(-2)-12) = (-2,-16).

Q: What are the x-intercepts of y=x^2+4x-12?

Set the expression to zero: x^2+4x-12=0. Using (x+6)(x-2)=0, the intercepts are (-6,0) and (2,0).

Q: What is the y-intercept of y=x^2+4x-12?

Substitute x=0: y=0^2+4\cdot0-12=-12. So the y-intercept is (0,-12).

Q: How do you use the quadratic formula on x^2+4x-12=0?

a=1,b=4,c=-12. x=\frac{-4\pm\sqrt{4^2-4(1)(-12)}}{2}=\frac{-4\pm\sqrt{64}}{2}=\frac{-4\pm8}{2}, so x=2 or x=-6.

Answer

\(x^2+4x-12\) is a quadratic. Factor it:

\[
x^2+4x-12=(x+6)(x-2)
\]

Set each factor equal to zero:

\[
x+6=0 \Rightarrow x=-6,\quad x-2=0 \Rightarrow x=2
\]

Final result: \(x=-6\) or \(x=2\).

Detailed Explanation

We want to factor (or simplify) the expression \(x^2+4x-12\).

Step 1: Identify the coefficients

Write the expression in the standard quadratic form \(ax^2+bx+c\).

\(x^2+4x-12\) corresponds to:

\(a=1\)
\(b=4\)
\(c=-12\)

Step 2: Use the factoring method

Because \(a=1\), we look for two numbers that multiply to \(c\) and add to \(b\).

So we need numbers \(m\) and \(n\) such that:

\(m\cdot n=-12\) and \(m+n=4\).

Step 3: Find the correct pair

Check factor pairs of \(-12\):

\(3\) and \(-4\) give \(3+(-4)=-1\)
\(-3\) and \(4\) give \(-3+4=1\)
Actually, the pair that works for \(-12\) and \(4\) is \(6\) and \(-2\) because \(6\cdot(-2)=-12\) and \(6+(-2)=4\)

Step 4: Write the factored form

Use the two numbers found: \(6\) and \(-2\).

\(x^2+4x-12=(x+6)(x-2)\).

Final Answer

\[
x^2+4x-12=(x+6)(x-2)
\]

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Algebra FAQ

What are the solutions to \(x^2+4x-12=0\)?

Factor \(x^2+4x-12=(x+6)(x-2)\). So \(x=-6\) or \(x=2\).

Can you complete the square for \(x^2+4x-12\)?

\(x^2+4x-12=(x+2)^2-16\). Then \((x+2)^2=16\), giving \(x=2\) or \(x=-6\).

What is the vertex of \(y=x^2+4x-12\)?

For \(y=x^2+4x-12\), \(x\)-coordinate is \(-\frac{b}{2a}=-\frac{4}{2}=-2\). Vertex is at \((-2,( -2)^2+4(-2)-12) = (-2,-16)\).

What are the \(x\)-intercepts of \(y=x^2+4x-12\)?

Set the expression to zero: \(x^2+4x-12=0\). Using \((x+6)(x-2)=0\), the intercepts are \((-6,0)\) and \((2,0)\).

What is the \(y\)-intercept of \(y=x^2+4x-12\)?

Substitute \(x=0\): \(y=0^2+4\cdot0-12=-12\). So the \(y\)-intercept is \((0,-12)\).

How do you use the quadratic formula on \(x^2+4x-12=0\)?

\(a=1,b=4,c=-12\). \(x=\frac{-4\pm\sqrt{4^2-4(1)(-12)}}{2}=\frac{-4\pm\sqrt{64}}{2}=\frac{-4\pm8}{2}\), so \(x=2\) or \(x=-6\).

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