Question

ving completely. notice that there are no equal signs in the expression ers should be factors, not single numbers.
6.) $3x^2 - 5x - 8$
8.) $-\frac{1}{2}x^2 + 6x - 10$
write any non-integer answers in exact a

Explanation:

Problem 6: Factoring \( 3x^2 - 5x - 8 \)

Step 1: Find two numbers

We need two numbers that multiply to \( 3 \times (-8) = -24 \) and add up to \( -5 \). The numbers are \( -8 \) and \( 3 \) because \( -8 \times 3 = -24 \) and \( -8 + 3 = -5 \).

Step 2: Split the middle term

Rewrite the middle term using these two numbers: \( 3x^2 + 3x - 8x - 8 \).

Step 3: Group and factor

Group the first two terms and the last two terms: \( (3x^2 + 3x) + (-8x - 8) \). Factor out the greatest common factor from each group: \( 3x(x + 1) - 8(x + 1) \). Then factor out the common binomial factor \( (x + 1) \): \( (3x - 8)(x + 1) \).

Step 1: Factor out the common factor

First, factor out \( -\frac{1}{2} \) from all terms: \( -\frac{1}{2}(x^2 - 12x + 20) \).

Step 2: Factor the quadratic inside the parentheses

We need two numbers that multiply to \( 20 \) and add up to \( -12 \). The numbers are \( -10 \) and \( -2 \) because \( -10 \times -2 = 20 \) and \( -10 + (-2) = -12 \). Rewrite the middle term: \( x^2 - 10x - 2x + 20 \).

Step 3: Group and factor

Group the first two terms and the last two terms: \( (x^2 - 10x) + (-2x + 20) \). Factor out the greatest common factor from each group: \( x(x - 10) - 2(x - 10) \). Then factor out the common binomial factor \( (x - 10) \): \( (x - 10)(x - 2) \).

Step 4: Combine with the factored out term

Put it all together with the \( -\frac{1}{2} \) we factored out earlier: \( -\frac{1}{2}(x - 10)(x - 2) \).

Answer:

\( (3x - 8)(x + 1) \)

Problem 8: Factoring \( -\frac{1}{2}x^2 + 6x - 10 \)