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\left(x-7\right)\left(x^{2}+4x-5\right)
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 35 and q divides the leading coefficient 1. One such root is 7. Factor the polynomial by dividing it by x-7.
a+b=4 ab=1\left(-5\right)=-5
Consider x^{2}+4x-5. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.
a=-1 b=5
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(5x-5\right)
Rewrite x^{2}+4x-5 as \left(x^{2}-x\right)+\left(5x-5\right).
x\left(x-1\right)+5\left(x-1\right)
Factor out x in the first and 5 in the second group.
\left(x-1\right)\left(x+5\right)
Factor out common term x-1 by using distributive property.
\left(x-7\right)\left(x-1\right)\left(x+5\right)
Rewrite the complete factored expression.