Solve 3p^4+28p^3+60p^2

$$3 p ^ { 4 } + 28 p ^ { 3 } + 60 p ^ { 2 }$$

Factor

$$\left(p+6\right)\left(3p+10\right)p^{2}$$

Solution Steps

Factor out p^{2}.

$$p^{2}\left(3p^{2}+28p+60\right)$$

Consider 3p^{2}+28p+60. Factor the expression by grouping. First, the expression needs to be rewritten as 3p^{2}+ap+bp+60. To find a and b, set up a system to be solved.

$$a+b=28 ab=3\times 60=180$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 28.

$$a=10 b=18$$

Rewrite 3p^{2}+28p+60 as \left(3p^{2}+10p\right)+\left(18p+60\right).

$$\left(3p^{2}+10p\right)+\left(18p+60\right)$$

Factor out p in the first and 6 in the second group.

$$p\left(3p+10\right)+6\left(3p+10\right)$$

Factor out common term 3p+10 by using distributive property.

$$\left(3p+10\right)\left(p+6\right)$$

Rewrite the complete factored expression.

$$p^{2}\left(3p+10\right)\left(p+6\right)$$

Evaluate

$$\left(p+6\right)\left(3p+10\right)p^{2}$$