Solve 2n^3-18n^2+40n

$$2 n ^ { 3 } - 18 n ^ { 2 } + 40 n$$

Factor

$$2n\left(n-5\right)\left(n-4\right)$$

Solution Steps

Factor out 2.

$$2\left(n^{3}-9n^{2}+20n\right)$$

Consider n^{3}-9n^{2}+20n. Factor out n.

$$n\left(n^{2}-9n+20\right)$$

Consider n^{2}-9n+20. Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+20. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 20=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite n^{2}-9n+20 as \left(n^{2}-5n\right)+\left(-4n+20\right).

$$\left(n^{2}-5n\right)+\left(-4n+20\right)$$

Factor out n in the first and -4 in the second group.

$$n\left(n-5\right)-4\left(n-5\right)$$

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

$$\left(n-5\right)\left(n-4\right)$$

Rewrite the complete factored expression.

$$2n\left(n-5\right)\left(n-4\right)$$

Evaluate

$$2n\left(n-5\right)\left(n-4\right)$$