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\left(x+6\right)\left(x^{2}+3x-4\right)
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. One such root is -6. Factor the polynomial by dividing it by x+6.
a+b=3 ab=1\left(-4\right)=-4
Consider x^{2}+3x-4. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-4. To find a and b, set up a system to be solved.
-1,4 -2,2
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 -4.
-1+4=3 -2+2=0
Calculate the sum for each pair.
a=-1 b=4
The solution is the pair that gives sum 3.
\left(x^{2}-x\right)+\left(4x-4\right)
Rewrite x^{2}+3x-4 as \left(x^{2}-x\right)+\left(4x-4\right).
x\left(x-1\right)+4\left(x-1\right)
Factor out x in the first and 4 in the second group.
\left(x-1\right)\left(x+4\right)
Factor out common term x-1 by using distributive property.
\left(x-1\right)\left(x+4\right)\left(x+6\right)
Rewrite the complete factored expression.