Question

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2-3 lesson check: factored form of a quadratic function (d)
work must be shown for each problem. 3pts each.

1. factor the expression ( 5x^2 + x - 6 )
2. solve the equation ( x^2 + 14x = 32 ).
3. find the zeros of the function ( f(x) = x^2 - 11x + 18 ).

envision™ algebra 2 • assessment resources

Explanation:

Problem 1: Factor \( 5x^2 + x - 6 \)

Step 1: Find \( ac \) and split middle term

For \( ax^2 + bx + c \), \( a = 5 \), \( b = 1 \), \( c = -6 \).
\( ac = 5 \times (-6) = -30 \). Find two numbers that multiply to \( -30 \) and add to \( 1 \): \( 6 \) and \( -5 \).
Rewrite middle term: \( 5x^2 + 6x - 5x - 6 \).

Step 2: Group and factor

Group: \( (5x^2 + 6x) + (-5x - 6) \).
Factor: \( x(5x + 6) - 1(5x + 6) \).
Factor out \( (5x + 6) \): \( (5x + 6)(x - 1) \)? Wait, no—wait, original middle term after split: \( 5x^2 + 6x - 5x - 6 \). Wait, correction: \( 5x^2 + 6x - 5x - 6 = x(5x + 6) - 1(5x + 6) = (5x + 6)(x - 1) \)? But the handwritten answer was \( (3x + 2)(x - 3) \), which is wrong. Let's redo:

Wait, correct split: \( 5x^2 + x - 6 \). \( ac = -30 \), numbers: \( 6 \) and \( -5 \) (since \( 6 + (-5) = 1 \)). So:
\( 5x^2 + 6x - 5x - 6 = x(5x + 6) - 1(5x + 6) = (5x + 6)(x - 1) \)? No, that's incorrect. Wait, maybe I made a mistake. Let's use trial factors.
\( 5x^2 + x - 6 \). Possible factors: \( (5x + a)(x + b) \), \( ab = -6 \), \( 5b + a = 1 \).
Try \( a = -6 \), \( b = 1 \): \( 5(1) + (-6) = -1 \) no.
\( a = 6 \), \( b = -1 \): \( 5(-1) + 6 = 1 \). Yes! So \( (5x + 6)(x - 1) \)? Wait, no: \( (5x + 6)(x - 1) = 5x^2 - 5x + 6x - 6 = 5x^2 + x - 6 \). Correct. The handwritten answer \( (3x + 2)(x - 3) = 3x^2 - 9x + 2x - 6 = 3x^2 - 7x - 6 \), which is wrong.

Step 1: Rewrite as standard quadratic

Subtract 32: \( x^2 + 14x - 32 = 0 \).

Step 2: Factor (or use quadratic formula)

Find two numbers: multiply to \( -32 \), add to \( 14 \): \( 16 \) and \( -2 \).
Factor: \( (x + 16)(x - 2) = 0 \).

Step 3: Solve for \( x \)

Set each factor to 0:
\( x + 16 = 0 \implies x = -16 \)
\( x - 2 = 0 \implies x = 2 \)

Step 1: Factor the quadratic

Find two numbers: multiply to \( 18 \), add to \( -11 \): \( -9 \) and \( -2 \).
Factor: \( (x - 9)(x - 2) = 0 \).

Step 2: Solve for \( x \)

Set each factor to 0:
\( x - 9 = 0 \implies x = 9 \)
\( x - 2 = 0 \implies x = 2 \)

Answer:

(correct): \( (5x + 6)(x - 1) \)

Problem 2: Solve \( x^2 + 14x = 32 \)