Factor
\left(x-1\right)\left(5x+12\right)
Evaluate
\left(x-1\right)\left(5x+12\right)
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a+b=7 ab=5\left(-12\right)=-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 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 -60.
-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4
Calculate the sum for each pair.
a=-5 b=12
The solution is the pair that gives sum 7.
\left(5x^{2}-5x\right)+\left(12x-12\right)
Rewrite 5x^{2}+7x-12 as \left(5x^{2}-5x\right)+\left(12x-12\right).
5x\left(x-1\right)+12\left(x-1\right)
Factor out 5x in the first and 12 in the second group.
\left(x-1\right)\left(5x+12\right)
Factor out common term x-1 by using distributive property.
5x^{2}+7x-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{-7±\sqrt{7^{2}-4\times 5\left(-12\right)}}{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{-7±\sqrt{49-4\times 5\left(-12\right)}}{2\times 5}
Square 7.
x=\frac{-7±\sqrt{49-20\left(-12\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-7±\sqrt{49+240}}{2\times 5}
Multiply -20 times -12.
x=\frac{-7±\sqrt{289}}{2\times 5}
Add 49 to 240.
x=\frac{-7±17}{2\times 5}
Take the square root of 289.
x=\frac{-7±17}{10}
Multiply 2 times 5.
x=\frac{10}{10}
Now solve the equation x=\frac{-7±17}{10} when ± is plus. Add -7 to 17.
x=1
Divide 10 by 10.
x=-\frac{24}{10}
Now solve the equation x=\frac{-7±17}{10} when ± is minus. Subtract 17 from -7.
x=-\frac{12}{5}
Reduce the fraction \frac{-24}{10} to lowest terms by extracting and canceling out 2.
5x^{2}+7x-12=5\left(x-1\right)\left(x-\left(-\frac{12}{5}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -\frac{12}{5} for x_{2}.
5x^{2}+7x-12=5\left(x-1\right)\left(x+\frac{12}{5}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
5x^{2}+7x-12=5\left(x-1\right)\times \frac{5x+12}{5}
Add \frac{12}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
5x^{2}+7x-12=\left(x-1\right)\left(5x+12\right)
Cancel out 5, the greatest common factor in 5 and 5.
x ^ 2 +\frac{7}{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{7}{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{7}{10} - u s = -\frac{7}{10} + u
Two numbers r and s sum up to -\frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{5} = -\frac{7}{10}. 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}{10} - u) (-\frac{7}{10} + u) = -\frac{12}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{12}{5}
\frac{49}{100} - u^2 = -\frac{12}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{12}{5}-\frac{49}{100} = -\frac{289}{100}
Simplify the expression by subtracting \frac{49}{100} on both sides
u^2 = \frac{289}{100} u = \pm\sqrt{\frac{289}{100}} = \pm \frac{17}{10}
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}{10} - \frac{17}{10} = -2.400 s = -\frac{7}{10} + \frac{17}{10} = 1
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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