The expression $10 x^{2} - x - 24$ $10x^2-x-24$ can be written as $(A x - 8) (B x + 3)$ $(Ax-8)(Bx+3)$ , where $A$ $A$ and $B$ $B$ are integers. What is $A B + B$ $AB + B$ ?

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Answer 1 · 2018-06-07T00:25:49

$A B + B = 12, \frac{65}{8}$ $AB+B=12, 65/8$

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$10 x^{2} - x - 24 = (A x - 8) (B x + 3)$ $10x^2-x-24=(Ax-8)(Bx+3)$

$10 x^{2} - x - 24 = A B x^{2} + 3 A x - 8 B x - 24$ $10x^2-x-24=ABx^2+3Ax-8Bx-24$

$10 x^{2} - x - 24 = A B x^{2} - (8 B - 3 A) x - 24$ $10x^2-x-24=ABx^2-(8B-3A)x-24$

$A B = 10$ $AB=10$

$8B-3A=1, :. 8B=1+3A, B=(1+3A)/8$

$(A(1+3A))/8=10$

$3A^2+A-80=0$

$A=(-1+-sqrt(1-4(3)(-80)))/6=(-1+-sqrt961)/6=(-1+-31)/6$

$A=5, -16/3$

$A=5, :. B=2, :. AB+B=10+2=12$

$A=-16/3, :. B=-15/8, :. AB+B=10-15/8=65/8$