Question

1. quadratics:

factor ( x^2 - 2x - 15 )
find the zero’s ( x^2 + 5x - 14 )

Explanation:

Part 1: Factor \( x^2 - 2x - 15 \)

Step 1: Find two numbers

We need two numbers that multiply to \(-15\) (the constant term) and add up to \(-2\) (the coefficient of \(x\)). Let's list the factor pairs of \(-15\): \((-5, 3)\) because \(-5 \times 3 = -15\) and \(-5 + 3 = -2\).

Step 2: Rewrite and factor

Rewrite the middle term using these two numbers: \(x^2 - 5x + 3x - 15\). Then group the terms: \((x^2 - 5x) + (3x - 15)\). Factor out the greatest common factor from each group: \(x(x - 5) + 3(x - 5)\). Now, factor out the common binomial factor \((x - 5)\): \((x - 5)(x + 3)\).

Part 2: Find the zeros of \( x^2 + 5x - 14 \)

Step 1: Factor the quadratic

First, factor \(x^2 + 5x - 14\). We need two numbers that multiply to \(-14\) and add up to \(5\). The factor pair is \(7\) and \(-2\) because \(7 \times (-2) = -14\) and \(7 + (-2) = 5\). So, the factored form is \((x + 7)(x - 2)\).

Step 2: Set each factor to zero

To find the zeros, set each factor equal to zero:

• \(x + 7 = 0\) gives \(x = -7\)
• \(x - 2 = 0\) gives \(x = 2\)

Answer:

s:

• Factored form of \(x^2 - 2x - 15\): \(\boldsymbol{(x - 5)(x + 3)}\)
• Zeros of \(x^2 + 5x - 14\): \(\boldsymbol{x = -7}\) and \(\boldsymbol{x = 2}\)