Question

1. $(w^2 - 8w + 15) cdot \frac{w + 3}{4w - 20}$

Explanation:

Step1: Factor the quadratic and the linear expression

Factor \( w^2 - 8w + 15 \) using the formula \( x^2+(a + b)x+ab=(x + a)(x + b) \). We need two numbers that multiply to \( 15 \) and add to \( -8 \). The numbers are \( -3 \) and \( -5 \), so \( w^2 - 8w + 15=(w - 3)(w - 5) \).

Factor \( 4w - 20 \) by taking out the common factor \( 4 \), so \( 4w - 20 = 4(w - 5) \).

Now the expression becomes: \( (w - 3)(w - 5)\cdot\frac{w + 3}{4(w - 5)} \)

Step2: Cancel out the common factor

We can cancel out the common factor \( (w - 5) \) from the numerator and the denominator (assuming \( w
eq5 \) to avoid division by zero).

After canceling, we have: \( (w - 3)\cdot\frac{w + 3}{4} \)

Step3: Multiply the remaining factors

Multiply \( (w - 3) \) and \( (w + 3) \) using the difference of squares formula \( (a - b)(a + b)=a^2 - b^2 \), where \( a = w \) and \( b = 3 \). So \( (w - 3)(w + 3)=w^2 - 9 \).

Then the expression is \( \frac{(w - 3)(w + 3)}{4}=\frac{w^2 - 9}{4} \)

Answer:

\( \frac{w^2 - 9}{4} \) (or \( \frac{1}{4}w^2-\frac{9}{4} \))