Factor
\left(5b-3\right)\left(4b+5\right)
Evaluate
\left(5b-3\right)\left(4b+5\right)
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p+q=13 pq=20\left(-15\right)=-300
Factor the expression by grouping. First, the expression needs to be rewritten as 20b^{2}+pb+qb-15. To find p and q, set up a system to be solved.
-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20
Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -300.
-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5
Calculate the sum for each pair.
p=-12 q=25
The solution is the pair that gives sum 13.
\left(20b^{2}-12b\right)+\left(25b-15\right)
Rewrite 20b^{2}+13b-15 as \left(20b^{2}-12b\right)+\left(25b-15\right).
4b\left(5b-3\right)+5\left(5b-3\right)
Factor out 4b in the first and 5 in the second group.
\left(5b-3\right)\left(4b+5\right)
Factor out common term 5b-3 by using distributive property.
20b^{2}+13b-15=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.
b=\frac{-13±\sqrt{13^{2}-4\times 20\left(-15\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.
b=\frac{-13±\sqrt{169-4\times 20\left(-15\right)}}{2\times 20}
Square 13.
b=\frac{-13±\sqrt{169-80\left(-15\right)}}{2\times 20}
Multiply -4 times 20.
b=\frac{-13±\sqrt{169+1200}}{2\times 20}
Multiply -80 times -15.
b=\frac{-13±\sqrt{1369}}{2\times 20}
Add 169 to 1200.
b=\frac{-13±37}{2\times 20}
Take the square root of 1369.
b=\frac{-13±37}{40}
Multiply 2 times 20.
b=\frac{24}{40}
Now solve the equation b=\frac{-13±37}{40} when ± is plus. Add -13 to 37.
b=\frac{3}{5}
Reduce the fraction \frac{24}{40} to lowest terms by extracting and canceling out 8.
b=-\frac{50}{40}
Now solve the equation b=\frac{-13±37}{40} when ± is minus. Subtract 37 from -13.
b=-\frac{5}{4}
Reduce the fraction \frac{-50}{40} to lowest terms by extracting and canceling out 10.
20b^{2}+13b-15=20\left(b-\frac{3}{5}\right)\left(b-\left(-\frac{5}{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{3}{5} for x_{1} and -\frac{5}{4} for x_{2}.
20b^{2}+13b-15=20\left(b-\frac{3}{5}\right)\left(b+\frac{5}{4}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
20b^{2}+13b-15=20\times \frac{5b-3}{5}\left(b+\frac{5}{4}\right)
Subtract \frac{3}{5} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
20b^{2}+13b-15=20\times \frac{5b-3}{5}\times \frac{4b+5}{4}
Add \frac{5}{4} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
20b^{2}+13b-15=20\times \frac{\left(5b-3\right)\left(4b+5\right)}{5\times 4}
Multiply \frac{5b-3}{5} times \frac{4b+5}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
20b^{2}+13b-15=20\times \frac{\left(5b-3\right)\left(4b+5\right)}{20}
Multiply 5 times 4.
20b^{2}+13b-15=\left(5b-3\right)\left(4b+5\right)
Cancel out 20, the greatest common factor in 20 and 20.
x ^ 2 +\frac{13}{20}x -\frac{3}{4} = 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{13}{20} rs = -\frac{3}{4}
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{13}{40} - u s = -\frac{13}{40} + u
Two numbers r and s sum up to -\frac{13}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{20} = -\frac{13}{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{13}{40} - u) (-\frac{13}{40} + u) = -\frac{3}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}
\frac{169}{1600} - u^2 = -\frac{3}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{4}-\frac{169}{1600} = -\frac{1369}{1600}
Simplify the expression by subtracting \frac{169}{1600} on both sides
u^2 = \frac{1369}{1600} u = \pm\sqrt{\frac{1369}{1600}} = \pm \frac{37}{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{13}{40} - \frac{37}{40} = -1.250 s = -\frac{13}{40} + \frac{37}{40} = 0.600
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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