Solve 27-125a^3-135a+225a^2

$$27 - 125 a ^ { 3 } - 135 a + 225 a ^ { 2 }$$

Factor

$$\left(3-5a\right)^{3}$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 27 and q divides the leading coefficient -125. One such root is \frac{3}{5}. Factor the polynomial by dividing it by 5a-3.

$$\left(5a-3\right)\left(-25a^{2}+30a-9\right)$$

Consider -25a^{2}+30a-9. Factor the expression by grouping. First, the expression needs to be rewritten as -25a^{2}+pa+qa-9. To find p and q, set up a system to be solved.

$$p+q=30 pq=-25\left(-9\right)=225$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 225.

$$1,225 3,75 5,45 9,25 15,15$$

Calculate the sum for each pair.

$$1+225=226 3+75=78 5+45=50 9+25=34 15+15=30$$

The solution is the pair that gives sum 30.

$$p=15 q=15$$

Rewrite -25a^{2}+30a-9 as \left(-25a^{2}+15a\right)+\left(15a-9\right).

$$\left(-25a^{2}+15a\right)+\left(15a-9\right)$$

Factor out -5a in the first and 3 in the second group.

$$-5a\left(5a-3\right)+3\left(5a-3\right)$$

Factor out common term 5a-3 by using distributive property.

$$\left(5a-3\right)\left(-5a+3\right)$$

Rewrite the complete factored expression.

$$\left(-5a+3\right)\left(5a-3\right)^{2}$$

Evaluate

$$\left(3-5a\right)^{3}$$