Q. What is a solution to \( (x + 6)(x + 2) = 60 \)?

Answer

To find the solution for \((x+6)(x+2)=60\), follow these algebraic steps:

\[
\begin{aligned}
(x+6)(x+2)&=60\\
x^2+8x+12&=60\\
x^2+8x-48&=0\\
(x+12)(x-4)&=0\\
x&=-12 \text{ or } x=4
\end{aligned}
\]

Final result: \(\boxed{x=-12\text{ or }x=4}\).

Detailed Explanation

Problem: Solve the equation \( (x+6)(x+2)=60 \)

Step 1 — Expand the product

Use FOIL (distribute each term):

\[
(x+6)(x+2)=x\cdot x + x\cdot 2 + 6\cdot x + 6\cdot 2
= x^{2}+2x+6x+12
= x^{2}+8x+12
\]

So the equation becomes

\[
x^{2}+8x+12=60
\]

Step 2 — Move all terms to one side

\[
x^{2}+8x+12-60=0
\quad\Rightarrow\quad
x^{2}+8x-48=0
\]

Step 3 — Factor the quadratic

Find two numbers that multiply to \(-48\) and add to \(8\): these are \(12\) and \(-4\). Thus

\[
x^{2}+8x-48=(x+12)(x-4)
\]

Step 4 — Apply the zero-product principle

\[
(x+12)(x-4)=0
\]

\[
x+12=0 \quad \text{or} \quad x-4=0
\]

\[
x=-12 \quad \text{or} \quad x=4
\]

Check (optional)

\[
(4+6)(4+2)=10\cdot 6=60
\qquad
(-12+6)(-12+2)=(-6)\cdot(-10)=60
\]

Both solutions satisfy the original equation.

Final answer

\[
x=-12 \quad \text{or} \quad x=4
\]

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FAQs

What is the solution to ( (x + 6)(x + 2) = 60 )?

Expand to (x^2+8x+12=60), so (x^2+8x-48=0). Factor: ((x+12)(x-4)=0). Solutions: (x=-12) and (x=4).

How do I expand and rearrange this to standard quadratic form?

Multiply: ((x+6)(x+2)=x^2+8x+12). Subtract 60: (x^2+8x+12-60=0), giving (x^2+8x-48=0).

How can I factor (x^2+8x-48)?

Find two numbers with product (-48) and sum (8): (12) and (-4). So (x^2+8x-48=(x+12)(x-4)).

Can I use the quadratic formula and get the same answers?

Yes. For (x^2+8x-48=0), (x=frac{-8pmsqrt{8^2-4(1)(-48)}}{2}=frac{-8pmsqrt{64+192}}{2}=frac{-8pm16}{2}), giving (x=4) or (x=-12).

Are either of the solutions extraneous?

No. Both (x=4) and (x=-12) satisfy the original equation: ((4+6)(4+2)=60) and ((-12+6)(-12+2)=60).

What are the sum and product of the roots?

For (x^2+8x-48=0), sum = (-8) (because (-b/a=-8)), product = (-48) (c/a). Indeed (4+(-12)=-8) and (4cdot(-12)=-48).

How can I check the solutions quickly?

Substitute into the original: for (x=4), ((10)(6)=60); for (x=-12), ((-6)(-10)=60). Both check.

What is the graphical interpretation of these solutions?

They are x-coordinates where the parabola (y=(x+6)(x+2)) intersects the horizontal line (y=60). The intersections occur at (x=-12) and (x=4).

Could I solve this by completing the square?

Yes. From (x^2+8x-48=0), add 48: (x^2+8x=48). Complete square: ((x+4)^2-16=48), so ((x+4)^2=64), giving (x+4=pm8) and (x=4) or (x=-12).

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