Q. \(x^2 – 4x – 12\)

Q: What are the zeros of x^2-4x-12?

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

Q: Factor x^2-4x-12.

Look for numbers with product -12 and sum -4: -6 and 2. Thus (x-6)(x+2).

Q: Rewrite x^2-4x-12 in vertex form.

Vertex form is (x-2)^2-16, with vertex (2,-16).

Q: What is the sign of x^2-4x-12 across real x?

Since it factors as (x-6)(x+2), it is positive for x 6, and negative for -2<x<6.

Answer

We factor the quadratic:

\[
x^2 – 4x – 12 = (x-6)(x+2)
\]

Set each factor 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 and/or solve the expression \(x^2 – 4x – 12\). A common goal with a quadratic like this is to factor it into a product of two binomials.

Step 1: Identify the quadratic form

Write the expression as:

\[
x^2 – 4x – 12
\]

Here, we have \(a = 1\), \(b = -4\), and \(c = -12\).

Step 2: Use the factoring idea

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

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

Step 3: Find numbers that multiply to \(-12\)

List factor pairs of \(-12\):

\(1\) and \(-12\) (sum \(= -11\))
\(-1\) and \(12\) (sum \(= 11\))
\(2\) and \(-6\) (sum \(= -4\))
\(-2\) and \(6\) (sum \(= 4\))

The pair that works is \(2\) and \(-6\), because:

\(2 \cdot (-6) = -12\)
\(2 + (-6) = -4\)

Step 4: Write the factored form

Since the correct numbers are \(2\) and \(-6\), the quadratic factors as:

\[
x^2 – 4x – 12 = (x + 2)(x – 6)
\]

Final Answer

\[
x^2 – 4x – 12 = (x + 2)(x – 6)
\]

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Algebra FAQ

What are the zeros of \(x^2-4x-12\)?

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

Factor \(x^2-4x-12\).

Look for numbers with product \(-12\) and sum \(-4\): \(-6\) and \(2\). Thus \((x-6)(x+2)\).

Solve \(x^2-4x-12=0\) using the quadratic formula.

\(x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(-12)}}{2}=\dfrac{4\pm\sqrt{64}}{2}=\dfrac{4\pm 8}{2}\), so \(x=6,-2\).

Complete the square for \(x^2-4x-12\).

\(x^2-4x-12=(x-2)^2-16\).

Rewrite \(x^2-4x-12\) in vertex form.

Vertex form is \((x-2)^2-16\), with vertex \((2,-16)\).

What is the sign of \(x^2-4x-12\) across real \(x\)?

Since it factors as \((x-6)(x+2)\), it is positive for \(x<-2\) and \(x>6\), and negative for \(-2

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