Question

jason needs to find the zeros of $y = 3x^2 - 5x + 2$. which factoring will help him determine the zeros?
$y = x(3x - 5) + 2$
$y = (3x - 2)(x - 1)$
$y = (3x + 2)(x - 1)$
$y = (3x - 1)(x - 2)$

Explanation:

Step1: Recall factoring quadratic

To find zeros of \( y = 3x^2 - 5x + 2 \), we factor the quadratic. We need two numbers that multiply to \( 3\times2 = 6 \) and add to \( -5 \). The numbers are \( -2 \) and \( -3 \).

Step2: Rewrite and factor

Rewrite the middle term: \( 3x^2 - 3x - 2x + 2 \). Group: \( (3x^2 - 3x) + (-2x + 2) = 3x(x - 1) - 2(x - 1) = (3x - 2)(x - 1) \). So \( y=(3x - 2)(x - 1) \).

Step3: Check other options

• First option: Expand \( x(3x - 5)+2 = 3x^2 - 5x + 2 \), but it's not factored.
• Third option: Expand \( (3x + 2)(x - 1)=3x^2 - x - 2

eq 3x^2 - 5x + 2 \).

• Fourth option: Expand \( (3x - 1)(x - 2)=3x^2 - 7x + 2

eq 3x^2 - 5x + 2 \).

Answer:

\( y=(3x - 2)(x - 1) \) (the second option)