Skip to main content
Factor
Tick mark Image
Evaluate
Tick mark Image
Graph

Similar Problems from Web Search

Share

a+b=-16 ab=5\times 12=60
Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+12. To find a and b, set up a system to be solved.
-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.
-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16
Calculate the sum for each pair.
a=-10 b=-6
The solution is the pair that gives sum -16.
\left(5x^{2}-10x\right)+\left(-6x+12\right)
Rewrite 5x^{2}-16x+12 as \left(5x^{2}-10x\right)+\left(-6x+12\right).
5x\left(x-2\right)-6\left(x-2\right)
Factor out 5x in the first and -6 in the second group.
\left(x-2\right)\left(5x-6\right)
Factor out common term x-2 by using distributive property.
5x^{2}-16x+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(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 12}}{2\times 5}
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(-16\right)±\sqrt{256-4\times 5\times 12}}{2\times 5}
Square -16.
x=\frac{-\left(-16\right)±\sqrt{256-20\times 12}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 5}
Multiply -20 times 12.
x=\frac{-\left(-16\right)±\sqrt{16}}{2\times 5}
Add 256 to -240.
x=\frac{-\left(-16\right)±4}{2\times 5}
Take the square root of 16.
x=\frac{16±4}{2\times 5}
The opposite of -16 is 16.
x=\frac{16±4}{10}
Multiply 2 times 5.
x=\frac{20}{10}
Now solve the equation x=\frac{16±4}{10} when ± is plus. Add 16 to 4.
x=2
Divide 20 by 10.
x=\frac{12}{10}
Now solve the equation x=\frac{16±4}{10} when ± is minus. Subtract 4 from 16.
x=\frac{6}{5}
Reduce the fraction \frac{12}{10} to lowest terms by extracting and canceling out 2.
5x^{2}-16x+12=5\left(x-2\right)\left(x-\frac{6}{5}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and \frac{6}{5} for x_{2}.
5x^{2}-16x+12=5\left(x-2\right)\times \frac{5x-6}{5}
Subtract \frac{6}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
5x^{2}-16x+12=\left(x-2\right)\left(5x-6\right)
Cancel out 5, the greatest common factor in 5 and 5.
x ^ 2 -\frac{16}{5}x +\frac{12}{5} = 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 5
r + s = \frac{16}{5} rs = \frac{12}{5}
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{8}{5} - u s = \frac{8}{5} + u
Two numbers r and s sum up to \frac{16}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{5} = \frac{8}{5}. 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{8}{5} - u) (\frac{8}{5} + u) = \frac{12}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{12}{5}
\frac{64}{25} - u^2 = \frac{12}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{12}{5}-\frac{64}{25} = -\frac{4}{25}
Simplify the expression by subtracting \frac{64}{25} on both sides
u^2 = \frac{4}{25} u = \pm\sqrt{\frac{4}{25}} = \pm \frac{2}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{8}{5} - \frac{2}{5} = 1.200 s = \frac{8}{5} + \frac{2}{5} = 2.000
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.