Question

can you factor the trinomial (6x^2 + 7x - 6) as ((px + q)(sx + t)), where (p), (q), (s), and (t) are integers? explain why or why not. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. yes. the trinomial can be factored as (square). b. no. the trinomial cannot be factored into that form because there are no factors of (-36) that sum to (7).

Explanation:

Step1: Multiplicar \( p \) y \( s \)

Para factorizar \( 6x^2 + 7x - 6 \) en la forma \( (px + q)(sx + t) \), primero, \( p \times s = 6 \). Las posibles parejas de enteros son \( (1,6) \), \( (2,3) \), \( (3,2) \), \( (6,1) \).

Step2: Encontrar \( q \) y \( t \) tales que \( q \times t = -6 \) y \( p \times t + s \times q = 7 \)

Probemos \( p = 2 \), \( s = 3 \). Entonces \( q \times t = -6 \). Probemos \( q = -1 \), \( t = 6 \): \( 2 \times 6 + 3 \times (-1) = 12 - 3 = 9 eq 7 \). Probemos \( q = -2 \), \( t = 3 \): \( 2 \times 3 + 3 \times (-2) = 6 - 6 = 0 eq 7 \). Probemos \( q = 3 \), \( t = -2 \): \( 2 \times (-2) + 3 \times 3 = -4 + 9 = 5 eq 7 \). Probemos \( q = 6 \), \( t = -1 \): \( 2 \times (-1) + 3 \times 6 = -2 + 18 = 16 eq 7 \).

Ahora probemos \( p = 3 \), \( s = 2 \). \( q \times t = -6 \). Probemos \( q = -1 \), \( t = 6 \): \( 3 \times 6 + 2 \times (-1) = 18 - 2 = 16 eq 7 \). Probemos \( q = -2 \), \( t = 3 \): \( 3 \times 3 + 2 \times (-2) = 9 - 4 = 5 eq 7 \). Probemos \( q = 2 \), \( t = -3 \): \( 3 \times (-3) + 2 \times 2 = -9 + 4 = -5 eq 7 \). Probemos \( q = 3 \), \( t = -2 \): \( 3 \times (-2) + 2 \times 3 = -6 + 6 = 0 eq 7 \). Probemos \( q = -3 \), \( t = 2 \): \( 3 \times 2 + 2 \times (-3) = 6 - 6 = 0 eq 7 \). Probemos \( q = 6 \), \( t = -1 \): \( 3 \times (-1) + 2 \times 6 = -3 + 12 = 9 eq 7 \). Probemos \( q = -6 \), \( t = 1 \): \( 3 \times 1 + 2 \times (-6) = 3 - 12 = -9 eq 7 \).

Ahora probemos \( p = 1 \), \( s = 6 \). \( q \times t = -6 \). Probemos \( q = -1 \), \( t = 6 \): \( 1 \times 6 + 6 \times (-1) = 6 - 6 = 0 eq 7 \). Probemos \( q = -2 \), \( t = 3 \): \( 1 \times 3 + 6 \times (-2) = 3 - 12 = -9 eq 7 \). Probemos \( q = 3 \), \( t = -2 \): \( 1 \times (-2) + 6 \times 3 = -2 + 18 = 16 eq 7 \). Probemos \( q = 6 \), \( t = -1 \): \( 1 \times (-1) + 6 \times 6 = -1 + 36 = 35 eq 7 \).

Ahora probemos \( p = 6 \), \( s = 1 \). \( q \times t = -6 \). Probemos \( q = -1 \), \( t = 6 \): \( 6 \times 6 + 1 \times (-1) = 36 - 1 = 35 eq 7 \). Probemos \( q = -2 \), \( t = 3 \): \( 6 \times 3 + 1 \times (-2) = 18 - 2 = 16 eq 7 \). Probemos \( q = 2 \), \( t = -3 \): \( 6 \times (-3) + 1 \times 2 = -18 + 2 = -16 eq 7 \). Probemos \( q = 3 \), \( t = -2 \): \( 6 \times (-2) + 1 \times 3 = -12 + 3 = -9 eq 7 \). Probemos \( q = -3 \), \( t = 2 \): \( 6 \times 2 + 1 \times (-3) = 12 - 3 = 9 eq 7 \). Probemos \( q = 6 \), \( t = -1 \): \( 6 \times (-1) + 1 \times 6 = -6 + 6 = 0 eq 7 \).

Wait, maybe I made a mistake. Let's use the formula for factoring \( ax^2 + bx + c \). We need two numbers that multiply to \( a \times c = 6 \times (-6) = -36 \) and add up to \( b = 7 \). Let's find two numbers that multiply to -36 and add to 7. The numbers are 12 and -3, because \( 12 \times (-3) = -36 \) and \( 12 + (-3) = 9 \)? No, 9 is not 7. Wait, 9 and -4: \( 9 \times (-4) = -36 \), \( 9 + (-4) = 5 \). 18 and -2: \( 18 \times (-2) = -36 \), \( 18 + (-2) = 16 \). 36 and -1: \( 36 + (-1) = 35 \). Wait, maybe there's a mistake. Wait, the trinomial is \( 6x^2 + 7x - 6 \). Let's try again.

Wait, let's use the method of splitting the middle term. \( 6x^2 + 7x - 6 \). We need two numbers that multiply to \( 6 \times (-6) = -36 \) and add to 7. Let's list the factors of -36: (1, -36), (2, -18), (3, -12), (4, -9), (6, -6), (9, -4), (12, -3), (18, -2), (36, -1), and their negatives. Now, which pair adds to 7? Let's check: 12 and -3: 12 + (-3) = 9. 9 and -4: 5. 18 and -2: 16. 36 and -1: 35. 4 and -9: -5. 3 and -12: -9. 2 and -18: -16. 1 and -36: -35. Wait, none of these add to 7. Wait, but that can't be. Wait, maybe I miscalculated. Wait, \( 6x^2 + 7x - 6 \). Let's try factoring:

Wait, let's try (2x + 3)(3x - 2): \( 2x \times 3x = 6x^2 \), \( 2x \times (-2) = -4x \), \( 3 \times 3x = 9x \), \( 3 \times (-2) = -6 \). Then combine like terms: \( -4x + 9x = 5x \). So \( (2x + 3)(3x - 2) = 6x^2 + 5x - 6 \). Not 7x.

Try (3x + 3)(2x - 2): No, that's 6x² -6x +6x -6 = 6x² -6. Not good.

Try (2x - 1)(3x + 6): \( 2x \times 3x = 6x² \), \( 2x \times 6 = 12x \), \( -1 \times 3x = -3x \), \( -1 \times 6 = -6 \). Combine: \( 12x - 3x = 9x \). So \( 6x² + 9x -6 \). Not 7x.

Try (3x - 1)(2x + 6): \( 3x \times 2x = 6x² \), \( 3x \times 6 = 18x \), \( -1 \times 2x = -2x \), \( -1 \times 6 = -6 \). Combine: \( 18x - 2x = 16x \). \( 6x² +16x -6 \). No.

Try (6x - 3)(x + 2): \( 6x \times x = 6x² \), \( 6x \times 2 = 12x \), \( -3 \times x = -3x \), \( -3 \times 2 = -6 \). Combine: \( 12x -3x = 9x \). \( 6x² +9x -6 \). No.

Try (6x + 3)(x - 2): \( 6x \times x = 6x² \), \( 6x \times (-2) = -12x \), \( 3 \times x = 3x \), \( 3 \times (-2) = -6 \). Combine: \( -12x +3x = -9x \). \( 6x² -9x -6 \). No.

Try (2x + 2)(3x - 3): \( 6x² -6x +6x -6 = 6x² -6 \). No.

Wait, maybe the trinomial cannot be factored into integers? But the options are A: Yes, factor as..., B: No, because no factors of -36 sum to 7. Wait, earlier when we tried to find two numbers that multiply to -36 and add to 7, we couldn't find any. So the correct answer is B? Wait, but let's check again.

Wait, the product of a and c is 6*(-6) = -36. We need two numbers that multiply to -36 and add to 7. Let's list all pairs:

1 and -36: sum 1 -36 = -35

2 and -18: 2 -18 = -16

3 and -12: 3 -12 = -9

4 and -9: 4 -9 = -5

6 and -6: 6 -6 = 0

9 and -4: 9 -4 = 5

12 and -3: 12 -3 = 9

18 and -2: 18 -2 = 16

36 and -1: 36 -1 = 35

None of these add up to 7. So there are no integers q and t such that p*s=6, q*t=-6, and p*t + s*q=7. Therefore, the trinomial cannot be factored into (px + q)(sx + t) with integer coefficients. So the correct option is B.

Answer:

B. No. The trinomial cannot be factored into that form because there are no factors of -36 that sum to 7.