Question

Difficulty: MediumFactoring Polynomials

When the polynomial 6x2−11x−106x^2 - 11x - 10 is factored into the form (ax+b)(cx+d)(ax + b)(cx + d), where aa, bb, cc, and dd are integers such that aa and cc are positive and a>ca > c, what is the value of ad−bcad - bc?

Answer: -19

Answer

The value of ad−bcad - bc is −19-19.
The correct answer is −19-19. Factoring the polynomial 6x2−11x−106x^2 - 11x - 10 yields (3x+2)(2x−5)(3x + 2)(2x - 5). Under the constraints that aa and cc are positive and a>ca > c, we must have a=3a = 3, b=2b = 2, c=2c = 2, and d=−5d = -5. Evaluating the expression ad−bcad - bc gives (3)(−5)−(2)(2)=−15−4=−19(3)(-5) - (2)(2) = -15 - 4 = -19.

Step-by-Step Solution

1
Factor the quadratic expression 6x2−11x−106x^2 - 11x - 10.
(3x+2)(2x−5)(3x + 2)(2x - 5)
Factoring by grouping is used to rewrite the quadratic trinomial.
2
Apply the positive coefficient constraints and a>ca > c to identify the constants.
a=3a = 3, b=2b = 2, c=2c = 2, d=−5d = -5
Since the leading coefficients must be positive and a>ca > c, we assign a=3a = 3 from the first factor and c=2c = 2 from the second factor.
3
Evaluate the expression ad−bcad - bc.
−19-19
Substitute the values of aa, bb, cc, and dd to calculate the final numerical value.

Key Concept

Factoring quadratic trinomials of the form ax2+bx+cax^2 + bx + c where a>1a > 1
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