Q. \(x^2 – 16\)

Answer

We factor the quadratic by finding numbers that multiply to \(-16\) and add to \(0\):

\[
x^2 – 16 = (x-4)(x+4)
\]

So the solutions come from each factor equal to \(0\):

\[
x-4=0 \Rightarrow x=4,\quad x+4=0 \Rightarrow x=-4
\]

Final result: \(x=4\) or \(x=-4\).

Detailed Explanation

We want to factor and/or rewrite the expression \(x^2 – 16\).

Step 1: Recognize a difference of squares

The form \(a^2 – b^2\) can be factored as \((a-b)(a+b)\).

Here, \(x^2 – 16\) matches \(a^2 – b^2\) with:

  • \(a = x\), since \(a^2 = x^2\)

Step 2: Apply the difference of squares formula

Using \((a-b)(a+b)\):

\[
x^2 – 16 = x^2 – 4^2 = (x-4)(x+4)
\]

Final Answer

\[
x^2 – 16 = (x-4)(x+4)
\]

See full solution

Graph

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Algebra FAQ

How do you factor \(x^2-16\)?

Use difference of squares: \(x^2-16=x^2-4^2=(x-4)(x+4)\).

What are the solutions to \(x^2-16=0\)?

Set \((x-4)(x+4)=0\). Then \(x=4\) or \(x=-4\).

Solve \(x^2-16\ge 0\).

From \((x-4)(x+4)\ge 0\), the product is nonnegative when \(x\le -4\) or \(x\ge 4\). Interval: \((-\infty,-4]\cup[4,\infty)\).

Solve \(x^2-16<0\).

\((x-4)(x+4)<0\) between roots, so \(-4

What is the domain and range of \(y=x^2-16\)?

Domain: all real numbers. Range: since \(x^2\ge 0\), \(y\ge -16\). So range is \([-16,\infty)\).

Find the vertex of \(y=x^2-16\).

It’s \(y=(x-0)^2-16\), so the vertex is \((0,-16)\).

Similar Questions

Solve \(y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1\). \(x^4-16\) \(x^7 + 28x^4 – 42x^3 + 6x + 11\). \( y^2 = 4 x^5 + 4 x^3 + x^2 + 4 x \). how to calculate \( \Delta h \) \(x^{7} – 21x^{4} + 35x^{2} – 6x + 18\).
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