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a+b=-2 ab=1\left(-15\right)=-15
Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-15. To find a and b, set up a system to be solved.
1,-15 3,-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 -15.
1-15=-14 3-5=-2
Calculate the sum for each pair.
a=-5 b=3
The solution is the pair that gives sum -2.
\left(t^{2}-5t\right)+\left(3t-15\right)
Rewrite t^{2}-2t-15 as \left(t^{2}-5t\right)+\left(3t-15\right).
t\left(t-5\right)+3\left(t-5\right)
Factor out t in the first and 3 in the second group.
\left(t-5\right)\left(t+3\right)
Factor out common term t-5 by using distributive property.
t^{2}-2t-15=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(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-15\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(-2\right)±\sqrt{4-4\left(-15\right)}}{2}
Square -2.
t=\frac{-\left(-2\right)±\sqrt{4+60}}{2}
Multiply -4 times -15.
t=\frac{-\left(-2\right)±\sqrt{64}}{2}
Add 4 to 60.
t=\frac{-\left(-2\right)±8}{2}
Take the square root of 64.
t=\frac{2±8}{2}
The opposite of -2 is 2.
t=\frac{10}{2}
Now solve the equation t=\frac{2±8}{2} when ± is plus. Add 2 to 8.
t=5
Divide 10 by 2.
t=-\frac{6}{2}
Now solve the equation t=\frac{2±8}{2} when ± is minus. Subtract 8 from 2.
t=-3
Divide -6 by 2.
t^{2}-2t-15=\left(t-5\right)\left(t-\left(-3\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 -3 for x_{2}.
t^{2}-2t-15=\left(t-5\right)\left(t+3\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 -2x -15 = 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 = 2 rs = -15
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 = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>
(1 - u) (1 + u) = -15
To solve for unknown quantity u, substitute these in the product equation rs = -15
1 - u^2 = -15
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -15-1 = -16
Simplify the expression by subtracting 1 on both sides
u^2 = 16 u = \pm\sqrt{16} = \pm 4
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - 4 = -3 s = 1 + 4 = 5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.