Factor
\left(5x-2\right)\left(4x+3\right)
Evaluate
\left(5x-2\right)\left(4x+3\right)
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a+b=7 ab=20\left(-6\right)=-120
Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-6. To find a and b, set up a system to be solved.
-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -120.
-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2
Calculate the sum for each pair.
a=-8 b=15
The solution is the pair that gives sum 7.
\left(20x^{2}-8x\right)+\left(15x-6\right)
Rewrite 20x^{2}+7x-6 as \left(20x^{2}-8x\right)+\left(15x-6\right).
4x\left(5x-2\right)+3\left(5x-2\right)
Factor out 4x in the first and 3 in the second group.
\left(5x-2\right)\left(4x+3\right)
Factor out common term 5x-2 by using distributive property.
20x^{2}+7x-6=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{-7±\sqrt{7^{2}-4\times 20\left(-6\right)}}{2\times 20}
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{-7±\sqrt{49-4\times 20\left(-6\right)}}{2\times 20}
Square 7.
x=\frac{-7±\sqrt{49-80\left(-6\right)}}{2\times 20}
Multiply -4 times 20.
x=\frac{-7±\sqrt{49+480}}{2\times 20}
Multiply -80 times -6.
x=\frac{-7±\sqrt{529}}{2\times 20}
Add 49 to 480.
x=\frac{-7±23}{2\times 20}
Take the square root of 529.
x=\frac{-7±23}{40}
Multiply 2 times 20.
x=\frac{16}{40}
Now solve the equation x=\frac{-7±23}{40} when ± is plus. Add -7 to 23.
x=\frac{2}{5}
Reduce the fraction \frac{16}{40} to lowest terms by extracting and canceling out 8.
x=-\frac{30}{40}
Now solve the equation x=\frac{-7±23}{40} when ± is minus. Subtract 23 from -7.
x=-\frac{3}{4}
Reduce the fraction \frac{-30}{40} to lowest terms by extracting and canceling out 10.
20x^{2}+7x-6=20\left(x-\frac{2}{5}\right)\left(x-\left(-\frac{3}{4}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and -\frac{3}{4} for x_{2}.
20x^{2}+7x-6=20\left(x-\frac{2}{5}\right)\left(x+\frac{3}{4}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
20x^{2}+7x-6=20\times \frac{5x-2}{5}\left(x+\frac{3}{4}\right)
Subtract \frac{2}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
20x^{2}+7x-6=20\times \frac{5x-2}{5}\times \frac{4x+3}{4}
Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
20x^{2}+7x-6=20\times \frac{\left(5x-2\right)\left(4x+3\right)}{5\times 4}
Multiply \frac{5x-2}{5} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
20x^{2}+7x-6=20\times \frac{\left(5x-2\right)\left(4x+3\right)}{20}
Multiply 5 times 4.
20x^{2}+7x-6=\left(5x-2\right)\left(4x+3\right)
Cancel out 20, the greatest common factor in 20 and 20.
x ^ 2 +\frac{7}{20}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 20
r + s = -\frac{7}{20} 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{7}{40} - u s = -\frac{7}{40} + u
Two numbers r and s sum up to -\frac{7}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{20} = -\frac{7}{40}. 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{7}{40} - u) (-\frac{7}{40} + u) = -\frac{3}{10}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}
\frac{49}{1600} - u^2 = -\frac{3}{10}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{10}-\frac{49}{1600} = -\frac{529}{1600}
Simplify the expression by subtracting \frac{49}{1600} on both sides
u^2 = \frac{529}{1600} u = \pm\sqrt{\frac{529}{1600}} = \pm \frac{23}{40}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{7}{40} - \frac{23}{40} = -0.750 s = -\frac{7}{40} + \frac{23}{40} = 0.400
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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