Factor
\left(5x-6\right)\left(2x+3\right)
Evaluate
\left(5x-6\right)\left(2x+3\right)
Graph
Share
Copied to clipboard
a+b=3 ab=10\left(-18\right)=-180
Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-18. To find a and b, set up a system to be solved.
-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15
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 -180.
-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3
Calculate the sum for each pair.
a=-12 b=15
The solution is the pair that gives sum 3.
\left(10x^{2}-12x\right)+\left(15x-18\right)
Rewrite 10x^{2}+3x-18 as \left(10x^{2}-12x\right)+\left(15x-18\right).
2x\left(5x-6\right)+3\left(5x-6\right)
Factor out 2x in the first and 3 in the second group.
\left(5x-6\right)\left(2x+3\right)
Factor out common term 5x-6 by using distributive property.
10x^{2}+3x-18=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{-3±\sqrt{3^{2}-4\times 10\left(-18\right)}}{2\times 10}
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{-3±\sqrt{9-4\times 10\left(-18\right)}}{2\times 10}
Square 3.
x=\frac{-3±\sqrt{9-40\left(-18\right)}}{2\times 10}
Multiply -4 times 10.
x=\frac{-3±\sqrt{9+720}}{2\times 10}
Multiply -40 times -18.
x=\frac{-3±\sqrt{729}}{2\times 10}
Add 9 to 720.
x=\frac{-3±27}{2\times 10}
Take the square root of 729.
x=\frac{-3±27}{20}
Multiply 2 times 10.
x=\frac{24}{20}
Now solve the equation x=\frac{-3±27}{20} when ± is plus. Add -3 to 27.
x=\frac{6}{5}
Reduce the fraction \frac{24}{20} to lowest terms by extracting and canceling out 4.
x=-\frac{30}{20}
Now solve the equation x=\frac{-3±27}{20} when ± is minus. Subtract 27 from -3.
x=-\frac{3}{2}
Reduce the fraction \frac{-30}{20} to lowest terms by extracting and canceling out 10.
10x^{2}+3x-18=10\left(x-\frac{6}{5}\right)\left(x-\left(-\frac{3}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{6}{5} for x_{1} and -\frac{3}{2} for x_{2}.
10x^{2}+3x-18=10\left(x-\frac{6}{5}\right)\left(x+\frac{3}{2}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
10x^{2}+3x-18=10\times \frac{5x-6}{5}\left(x+\frac{3}{2}\right)
Subtract \frac{6}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
10x^{2}+3x-18=10\times \frac{5x-6}{5}\times \frac{2x+3}{2}
Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
10x^{2}+3x-18=10\times \frac{\left(5x-6\right)\left(2x+3\right)}{5\times 2}
Multiply \frac{5x-6}{5} times \frac{2x+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
10x^{2}+3x-18=10\times \frac{\left(5x-6\right)\left(2x+3\right)}{10}
Multiply 5 times 2.
10x^{2}+3x-18=\left(5x-6\right)\left(2x+3\right)
Cancel out 10, the greatest common factor in 10 and 10.
x ^ 2 +\frac{3}{10}x -\frac{9}{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 10
r + s = -\frac{3}{10} rs = -\frac{9}{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{3}{20} - u s = -\frac{3}{20} + u
Two numbers r and s sum up to -\frac{3}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{10} = -\frac{3}{20}. 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{3}{20} - u) (-\frac{3}{20} + u) = -\frac{9}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{5}
\frac{9}{400} - u^2 = -\frac{9}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{9}{5}-\frac{9}{400} = -\frac{729}{400}
Simplify the expression by subtracting \frac{9}{400} on both sides
u^2 = \frac{729}{400} u = \pm\sqrt{\frac{729}{400}} = \pm \frac{27}{20}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{20} - \frac{27}{20} = -1.500 s = -\frac{3}{20} + \frac{27}{20} = 1.200
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}