Factor
\left(t-5\right)\left(t+4\right)
Evaluate
\left(t-5\right)\left(t+4\right)
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a+b=-1 ab=1\left(-20\right)=-20
Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-20. To find a and b, set up a system to be solved.
1,-20 2,-10 4,-5
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 -20.
1-20=-19 2-10=-8 4-5=-1
Calculate the sum for each pair.
a=-5 b=4
The solution is the pair that gives sum -1.
\left(t^{2}-5t\right)+\left(4t-20\right)
Rewrite t^{2}-t-20 as \left(t^{2}-5t\right)+\left(4t-20\right).
t\left(t-5\right)+4\left(t-5\right)
Factor out t in the first and 4 in the second group.
\left(t-5\right)\left(t+4\right)
Factor out common term t-5 by using distributive property.
t^{2}-t-20=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.
t=\frac{-\left(-1\right)±\sqrt{1-4\left(-20\right)}}{2}
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.
t=\frac{-\left(-1\right)±\sqrt{1+80}}{2}
Multiply -4 times -20.
t=\frac{-\left(-1\right)±\sqrt{81}}{2}
Add 1 to 80.
t=\frac{-\left(-1\right)±9}{2}
Take the square root of 81.
t=\frac{1±9}{2}
The opposite of -1 is 1.
t=\frac{10}{2}
Now solve the equation t=\frac{1±9}{2} when ± is plus. Add 1 to 9.
t=5
Divide 10 by 2.
t=-\frac{8}{2}
Now solve the equation t=\frac{1±9}{2} when ± is minus. Subtract 9 from 1.
t=-4
Divide -8 by 2.
t^{2}-t-20=\left(t-5\right)\left(t-\left(-4\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -4 for x_{2}.
t^{2}-t-20=\left(t-5\right)\left(t+4\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 -1x -20 = 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.
r + s = 1 rs = -20
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) = -20
To solve for unknown quantity u, substitute these in the product equation rs = -20
\frac{1}{4} - u^2 = -20
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -20-\frac{1}{4} = -\frac{81}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}
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} - \frac{9}{2} = -4 s = \frac{1}{2} + \frac{9}{2} = 5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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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}