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

$$\left. \begin{array} { l } { 27 - 125 a ^ { 3 } - 135 } \\ { + 225 a ^ { 2 } } \end{array} \right.$$

Factor

$$-\left(5a+3\right)\left(5a-6\right)^{2}$$

Solution Steps

Multiply and combine like terms.

$$-125a^{3}+225a^{2}-108$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -108 and q divides the leading coefficient -125. One such root is \frac{6}{5}. Factor the polynomial by dividing it by 5a-6.

$$\left(5a-6\right)\left(-25a^{2}+15a+18\right)$$

Consider -25a^{2}+15a+18. Factor the expression by grouping. First, the expression needs to be rewritten as -25a^{2}+pa+qa+18. To find p and q, set up a system to be solved.

$$p+q=15 pq=-25\times 18=-450$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -450.

$$-1,450 -2,225 -3,150 -5,90 -6,75 -9,50 -10,45 -15,30 -18,25$$

Calculate the sum for each pair.

$$-1+450=449 -2+225=223 -3+150=147 -5+90=85 -6+75=69 -9+50=41 -10+45=35 -15+30=15 -18+25=7$$

The solution is the pair that gives sum 15.

$$p=30 q=-15$$

Rewrite -25a^{2}+15a+18 as \left(-25a^{2}+30a\right)+\left(-15a+18\right).

$$\left(-25a^{2}+30a\right)+\left(-15a+18\right)$$

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

$$-5a\left(5a-6\right)-3\left(5a-6\right)$$

Factor out common term 5a-6 by using distributive property.

$$\left(5a-6\right)\left(-5a-3\right)$$

Rewrite the complete factored expression.

$$\left(-5a-3\right)\left(5a-6\right)^{2}$$

Evaluate

$$-\left(5a+3\right)\left(5a-6\right)^{2}$$

Solution Steps

Subtract 135 from 27 to get -108.

$$-108-125a^{3}+225a^{2}$$