Question

1. $2x^2 - 10x = 0$

$2x(x - 5)$
$2x = 0$ $x - 5 = 0$
$x = 0$ $x = 5$

1. $x^2 + 2x - 8 = 0$
2. $2x^2 - 32 = 0$

$2(x^2 - 16)$
$(x + 16)(x - 4)$

1. $(4x - 5)(4x - 5) = 0$

$x = \frac{5}{4}$

1. $6x^2 + x - 40 = 0$

extra scratch work

Explanation:

Let's solve each quadratic equation one by one:

Problem 1: \( 2x^2 - 10x = 0 \)

Step 1: Factor out the GCF

Factor out \( 2x \) from the left - hand side: \( 2x(x - 5)=0 \)

Step 2: Apply the zero - product property

If \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). So we have two equations:

• \( 2x=0 \), dividing both sides by 2 gives \( x = 0 \)
• \( x - 5=0 \), adding 5 to both sides gives \( x = 5 \)

Step 1: Factor the quadratic

We need to find two numbers that multiply to \( - 8 \) and add up to 2. The numbers are 4 and - 2. So we can factor the quadratic as \( (x + 4)(x - 2)=0 \)

Step 2: Apply the zero - product property

• If \( x + 4=0 \), then \( x=-4 \)
• If \( x - 2=0 \), then \( x = 2 \)

Step 1: Factor out the GCF

Factor out 2 from the left - hand side: \( 2(x^2 - 16)=0 \)

Step 2: Factor the difference of squares

Since \( x^2-16=x^2 - 4^2=(x + 4)(x - 4) \), the equation becomes \( 2(x + 4)(x - 4)=0 \)

Step 3: Apply the zero - product property

• \( x+4 = 0\Rightarrow x=-4 \)
• \( x - 4=0\Rightarrow x = 4 \) (Note: There was a mistake in the original scratch work where it was written as \( (x + 16)(x - 4) \), the correct factorization of \( x^2-16 \) is \( (x + 4)(x - 4) \))

Answer:

\( x = 0 \) or \( x = 5 \)

Problem 2: \( x^2+2x - 8 = 0 \)