Solve 20a^2+5a^3-15a^4

$$20 { a }^{ 2 } +5 { a }^{ 3 } -15 { a }^{ 4 }$$

Factor

$$-5\left(3a-4\right)\left(a+1\right)a^{2}$$

Solution Steps

Factor out 5.

$$5\left(4a^{2}+a^{3}-3a^{4}\right)$$

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

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

Consider 4+a-3a^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3a^{2}+a+4$$

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

$$p+q=1 pq=-3\times 4=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$p=4 q=-3$$

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

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

Factor out -a in the first and -1 in the second group.

$$-a\left(3a-4\right)-\left(3a-4\right)$$

Factor out common term 3a-4 by using distributive property.

$$\left(3a-4\right)\left(-a-1\right)$$

Rewrite the complete factored expression.

$$5a^{2}\left(3a-4\right)\left(-a-1\right)$$

Evaluate

$$-5\left(3a-4\right)\left(a+1\right)a^{2}$$