Question

name alyssa henner
13.3 practide math modeling - factoring trinomials ii (x² + bx + c)
date 10-18-23
section 13.3
insert the symbol + or - in each binomial factor to make the equation true.
1.) 3x² + 5x + 2 = (3x 2)(x 1)
2.) 3x² - x - 2 = (3x 2)(x 1)
fill in the blank in each binomial factor to make the equation true.
3.) 4x² + 11x + 6 = (4x + )( + 2)
4.) 4x² + 4x - 3 = (2x - )(2x + )
5.) 6x² + 11x - 7 = ( + 7)(2x - )
6.) 6x² - 53x + 40 = ( - 5)(x - )
factor the trinomial. if the trinomial cannot be factored, write prime. (ex. 2 on notes)

Explanation:

Problem 1: \( 3x^2 + 5x + 2 = (3x \_ 2)(x \_ 1) \)

Step 1: Multiply the binomials

Let's assume the symbols are \( + \) or \( - \). Multiply \( (3x + 2)(x + 1) \):

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Step 2: Check the result

The result matches the left - hand side of the equation \( 3x^2 + 5x + 2 \). So the symbols should be \( + \) for both blanks.

Step 1: Test the symbols

Let's try different combinations of \( + \) and \( - \). First, try \( (3x + 2)(x - 1) \):

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Step 2: Verify

The result is equal to the left - hand side \( 3x^2 - x - 2 \). So the first blank is \( + \) and the second blank is \( - \).

Step 1: Let the first blank be \( a \) and the second blank be \( bx + c \) (but we know the constant term of the second factor is 2)

We know that \( (4x + a)(bx + 2)=4x^2+11x + 6 \). Expanding \( (4x + a)(bx + 2)=4bx^2+(8 + ab)x+2a \)
We know that \( 4b = 4\), so \( b = 1 \). And \( 2a=6\), so \( a = 3 \). Let's check: \( (4x+3)(x + 2)=4x^2+8x+3x + 6=4x^2+11x + 6 \)

Step 2: Confirm

The expansion of \( (4x + 3)(x + 2) \) gives \( 4x^2+11x + 6 \), which matches the left - hand side.

Answer:

\( 3x^2 + 5x + 2=(3x + 2)(x + 1) \)

Problem 2: \( 3x^2 - x - 2=(3x \_ 2)(x \_ 1) \)