Factor
\left(5x-4\right)\left(8x+3\right)
Evaluate
\left(5x-4\right)\left(8x+3\right)
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a+b=-17 ab=40\left(-12\right)=-480
Factor the expression by grouping. First, the expression needs to be rewritten as 40x^{2}+ax+bx-12. To find a and b, set up a system to be solved.
1,-480 2,-240 3,-160 4,-120 5,-96 6,-80 8,-60 10,-48 12,-40 15,-32 16,-30 20,-24
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -480.
1-480=-479 2-240=-238 3-160=-157 4-120=-116 5-96=-91 6-80=-74 8-60=-52 10-48=-38 12-40=-28 15-32=-17 16-30=-14 20-24=-4
Calculate the sum for each pair.
a=-32 b=15
The solution is the pair that gives sum -17.
\left(40x^{2}-32x\right)+\left(15x-12\right)
Rewrite 40x^{2}-17x-12 as \left(40x^{2}-32x\right)+\left(15x-12\right).
8x\left(5x-4\right)+3\left(5x-4\right)
Factor out 8x in the first and 3 in the second group.
\left(5x-4\right)\left(8x+3\right)
Factor out common term 5x-4 by using distributive property.
40x^{2}-17x-12=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 40\left(-12\right)}}{2\times 40}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\left(-17\right)±\sqrt{289-4\times 40\left(-12\right)}}{2\times 40}
Square -17.
x=\frac{-\left(-17\right)±\sqrt{289-160\left(-12\right)}}{2\times 40}
Multiply -4 times 40.
x=\frac{-\left(-17\right)±\sqrt{289+1920}}{2\times 40}
Multiply -160 times -12.
x=\frac{-\left(-17\right)±\sqrt{2209}}{2\times 40}
Add 289 to 1920.
x=\frac{-\left(-17\right)±47}{2\times 40}
Take the square root of 2209.
x=\frac{17±47}{2\times 40}
The opposite of -17 is 17.
x=\frac{17±47}{80}
Multiply 2 times 40.
x=\frac{64}{80}
Now solve the equation x=\frac{17±47}{80} when ± is plus. Add 17 to 47.
x=\frac{4}{5}
Reduce the fraction \frac{64}{80} to lowest terms by extracting and canceling out 16.
x=-\frac{30}{80}
Now solve the equation x=\frac{17±47}{80} when ± is minus. Subtract 47 from 17.
x=-\frac{3}{8}
Reduce the fraction \frac{-30}{80} to lowest terms by extracting and canceling out 10.
40x^{2}-17x-12=40\left(x-\frac{4}{5}\right)\left(x-\left(-\frac{3}{8}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{5} for x_{1} and -\frac{3}{8} for x_{2}.
40x^{2}-17x-12=40\left(x-\frac{4}{5}\right)\left(x+\frac{3}{8}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
40x^{2}-17x-12=40\times \frac{5x-4}{5}\left(x+\frac{3}{8}\right)
Subtract \frac{4}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
40x^{2}-17x-12=40\times \frac{5x-4}{5}\times \frac{8x+3}{8}
Add \frac{3}{8} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
40x^{2}-17x-12=40\times \frac{\left(5x-4\right)\left(8x+3\right)}{5\times 8}
Multiply \frac{5x-4}{5} times \frac{8x+3}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
40x^{2}-17x-12=40\times \frac{\left(5x-4\right)\left(8x+3\right)}{40}
Multiply 5 times 8.
40x^{2}-17x-12=\left(5x-4\right)\left(8x+3\right)
Cancel out 40, the greatest common factor in 40 and 40.
x ^ 2 -\frac{17}{40}x -\frac{3}{10} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 40
r + s = \frac{17}{40} rs = -\frac{3}{10}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = \frac{17}{80} - u s = \frac{17}{80} + u
Two numbers r and s sum up to \frac{17}{40} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{40} = \frac{17}{80}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{17}{80} - u) (\frac{17}{80} + u) = -\frac{3}{10}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}
\frac{289}{6400} - u^2 = -\frac{3}{10}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{10}-\frac{289}{6400} = -\frac{2209}{6400}
Simplify the expression by subtracting \frac{289}{6400} on both sides
u^2 = \frac{2209}{6400} u = \pm\sqrt{\frac{2209}{6400}} = \pm \frac{47}{80}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{17}{80} - \frac{47}{80} = -0.375 s = \frac{17}{80} + \frac{47}{80} = 0.800
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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