Question

trinomials
7a) $10x^2 - 3x - 4$ (fix)
$(2x + 1)(5x - 4)$
light purple 8
a = 10
b = -3
c = -4
7.) $2x^2 - 5x + 3$
$(2x - 3)(x - 1)$
light green 2
a = 2
b = -5
c = 3
8.) $12x^2 - 13x - 3$
a = 12
b = -13
c = -3
9.) $6y^2 - x - 15$
a = 6
b = -1
c = -15
10.) $8x^2 - 13x + 6$
a = 8
b = -13
c = 6

Explanation:

Let's take problem 7: \(2x^2 - 5x + 3\) (assuming a typo, should be \(+3\) or \(-3\); let's use \(2x^2 - 5x + 3\) for factoring)

Step1: Identify \(a\), \(b\), \(c\)

For \(ax^2 + bx + c\), here \(a = 2\), \(b = -5\), \(c = 3\).

Step2: Find \(ac\)

\(ac = 2\times3 = 6\).

Step3: Find factors of \(ac\) that sum to \(b\)

Factors of \(6\) that sum to \(-5\): \(-2\) and \(-3\) (since \(-2 + (-3) = -5\)).

Step4: Split middle term

\(2x^2 - 2x - 3x + 3\).

Step5: Group and factor

\((2x^2 - 2x) + (-3x + 3) = 2x(x - 1) - 3(x - 1) = (2x - 3)(x - 1)\).

Answer:

For \(2x^2 - 5x + 3\), the factored form is \((2x - 3)(x - 1)\) (if the original was \(2x^2 - 5x - 3\), the process would adjust, but based on the note, this matches the given \((2x - 3)(x - 1)\)).