Factor
2\left(2x-3\right)\left(2x+1\right)
Evaluate
8x^{2}-8x-6
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2\left(4x^{2}-4x-3\right)
Factor out 2.
a+b=-4 ab=4\left(-3\right)=-12
Consider 4x^{2}-4x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
1,-12 2,-6 3,-4
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.
1-12=-11 2-6=-4 3-4=-1
Calculate the sum for each pair.
a=-6 b=2
The solution is the pair that gives sum -4.
\left(4x^{2}-6x\right)+\left(2x-3\right)
Rewrite 4x^{2}-4x-3 as \left(4x^{2}-6x\right)+\left(2x-3\right).
2x\left(2x-3\right)+2x-3
Factor out 2x in 4x^{2}-6x.
\left(2x-3\right)\left(2x+1\right)
Factor out common term 2x-3 by using distributive property.
2\left(2x-3\right)\left(2x+1\right)
Rewrite the complete factored expression.
8x^{2}-8x-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{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 8\left(-6\right)}}{2\times 8}
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(-8\right)±\sqrt{64-4\times 8\left(-6\right)}}{2\times 8}
Square -8.
x=\frac{-\left(-8\right)±\sqrt{64-32\left(-6\right)}}{2\times 8}
Multiply -4 times 8.
x=\frac{-\left(-8\right)±\sqrt{64+192}}{2\times 8}
Multiply -32 times -6.
x=\frac{-\left(-8\right)±\sqrt{256}}{2\times 8}
Add 64 to 192.
x=\frac{-\left(-8\right)±16}{2\times 8}
Take the square root of 256.
x=\frac{8±16}{2\times 8}
The opposite of -8 is 8.
x=\frac{8±16}{16}
Multiply 2 times 8.
x=\frac{24}{16}
Now solve the equation x=\frac{8±16}{16} when ± is plus. Add 8 to 16.
x=\frac{3}{2}
Reduce the fraction \frac{24}{16} to lowest terms by extracting and canceling out 8.
x=-\frac{8}{16}
Now solve the equation x=\frac{8±16}{16} when ± is minus. Subtract 16 from 8.
x=-\frac{1}{2}
Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.
8x^{2}-8x-6=8\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{1}{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{3}{2} for x_{1} and -\frac{1}{2} for x_{2}.
8x^{2}-8x-6=8\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
8x^{2}-8x-6=8\times \frac{2x-3}{2}\left(x+\frac{1}{2}\right)
Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
8x^{2}-8x-6=8\times \frac{2x-3}{2}\times \frac{2x+1}{2}
Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
8x^{2}-8x-6=8\times \frac{\left(2x-3\right)\left(2x+1\right)}{2\times 2}
Multiply \frac{2x-3}{2} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
8x^{2}-8x-6=8\times \frac{\left(2x-3\right)\left(2x+1\right)}{4}
Multiply 2 times 2.
8x^{2}-8x-6=2\left(2x-3\right)\left(2x+1\right)
Cancel out 4, the greatest common factor in 8 and 4.
x ^ 2 -1x -\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 8
r + s = 1 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{1}{2} - u s = \frac{1}{2} + u
Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{2}. 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{1}{2} - u) (\frac{1}{2} + u) = -\frac{3}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}
\frac{1}{4} - u^2 = -\frac{3}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{4}-\frac{1}{4} = -1
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = 1 u = \pm\sqrt{1} = \pm 1
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{2} - 1 = -0.500 s = \frac{1}{2} + 1 = 1.500
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}