Solve P_{1}(x)=x^3-7x^2+10x

$$P _ { 1 } ( x ) = x ^ { 3 } - 7 x ^ { 2 } + 10 x$$

Factor

$$x\left(x-5\right)\left(x-2\right)$$

Solution Steps

Factor out x.

$$x\left(x^{2}-7x+10\right)$$

Consider x^{2}-7x+10. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 10=10$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite x^{2}-7x+10 as \left(x^{2}-5x\right)+\left(-2x+10\right).

$$\left(x^{2}-5x\right)+\left(-2x+10\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-5\right)-2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-2\right)$$

Rewrite the complete factored expression.

$$x\left(x-5\right)\left(x-2\right)$$

Evaluate

$$x\left(x-5\right)\left(x-2\right)$$