Factor
\left(x-1\right)\left(x+3\right)\left(2x+5\right)
Evaluate
\left(x-1\right)\left(x+3\right)\left(2x+5\right)
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\left(2x+5\right)\left(x^{2}+2x-3\right)
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -15 and q divides the leading coefficient 2. One such root is -\frac{5}{2}. Factor the polynomial by dividing it by 2x+5.
a+b=2 ab=1\left(-3\right)=-3
Consider x^{2}+2x-3. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
a=-1 b=3
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. The only such pair is the system solution.
\left(x^{2}-x\right)+\left(3x-3\right)
Rewrite x^{2}+2x-3 as \left(x^{2}-x\right)+\left(3x-3\right).
x\left(x-1\right)+3\left(x-1\right)
Factor out x in the first and 3 in the second group.
\left(x-1\right)\left(x+3\right)
Factor out common term x-1 by using distributive property.
\left(x-1\right)\left(x+3\right)\left(2x+5\right)
Rewrite the complete factored expression.
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}