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\left(2x+1\right)\left(x^{3}-7x^{2}+7x+15\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{1}{2}. Factor the polynomial by dividing it by 2x+1.
\left(x-5\right)\left(x^{2}-2x-3\right)
Consider x^{3}-7x^{2}+7x+15. 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 1. One such root is 5. Factor the polynomial by dividing it by x-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=-3 b=1
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. The only such pair is the system solution.
\left(x^{2}-3x\right)+\left(x-3\right)
Rewrite x^{2}-2x-3 as \left(x^{2}-3x\right)+\left(x-3\right).
x\left(x-3\right)+x-3
Factor out x in x^{2}-3x.
\left(x-3\right)\left(x+1\right)
Factor out common term x-3 by using distributive property.
\left(x-5\right)\left(x-3\right)\left(x+1\right)\left(2x+1\right)
Rewrite the complete factored expression.