Solve -3ma^2+12ma-9m

$$- 3 m a ^ { 2 } + 12 m a - 9 m$$

Factor

$$3m\left(1-a\right)\left(a-3\right)$$

Solution Steps

Factor out 3.

$$3\left(-ma^{2}+4ma-3m\right)$$

Consider -ma^{2}+4ma-3m. Factor out m.

$$m\left(-a^{2}+4a-3\right)$$

Consider -a^{2}+4a-3. Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa-3. To find p and q, set up a system to be solved.

$$p+q=4 pq=-\left(-3\right)=3$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. The only such pair is the system solution.

$$p=3 q=1$$

Rewrite -a^{2}+4a-3 as \left(-a^{2}+3a\right)+\left(a-3\right).

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

Factor out -a in -a^{2}+3a.

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

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

$$\left(a-3\right)\left(-a+1\right)$$

Rewrite the complete factored expression.

$$3m\left(a-3\right)\left(-a+1\right)$$

Evaluate

$$3m\left(1-a\right)\left(a-3\right)$$