Question

factor. then list the x-intercepts. 1. $x^2 - 121$ 2. $25x^2 - 120x + 144$

Explanation:

For the first problem: \(x^2 - 121\)

Step1: Identify the form

The expression \(x^2 - 121\) is a difference of squares, since \(x^2=(x)^2\) and \(121 = 11^2\). The formula for factoring a difference of squares is \(a^2 - b^2=(a + b)(a - b)\).

Step2: Apply the difference of squares formula

Here, \(a = x\) and \(b = 11\). So, \(x^2-121=(x + 11)(x - 11)\).

Step3: Find the x - intercepts

To find the x - intercepts, we set the factored form equal to zero: \((x + 11)(x - 11)=0\).
Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So, \(x+11 = 0\) gives \(x=-11\) and \(x - 11=0\) gives \(x = 11\).

Step1: Identify the form

The expression \(25x^2-120x + 144\) is a perfect square trinomial. A perfect square trinomial has the form \(a^2-2ab + b^2=(a - b)^2\) (or \(a^2+2ab + b^2=(a + b)^2\)).
First, find \(a\) and \(b\) such that \((ax)^2=25x^2\) and \(b^2 = 144\). We have \(ax=\sqrt{25x^2}=5x\) (so \(a = 5\)) and \(b=\sqrt{144}=12\).
Now, check the middle term: \(2ab=2\times5x\times12 = 120x\), which matches the middle term of the trinomial (with a negative sign, so the form is \(a^2-2ab + b^2\)).

Step2: Apply the perfect square trinomial formula

Using the formula \(a^2-2ab + b^2=(a - b)^2\) with \(a = 5x\) and \(b = 12\), we get \(25x^2-120x + 144=(5x-12)^2\).

Step3: Find the x - intercepts

Set the factored form equal to zero: \((5x - 12)^2=0\).
Taking the square root of both sides, \(5x-12 = 0\). Solving for \(x\), we add 12 to both sides: \(5x=12\), then divide by 5: \(x=\frac{12}{5}=2.4\). Since the square of a number is zero only when the number itself is zero, this is a repeated root.

Answer:

Factored form: \((x + 11)(x - 11)\); x - intercepts: \(x=-11\) and \(x = 11\)

For the second problem: \(25x^2-120x + 144\)