Solve 40y^4+38y^3+6y^2

$$40 y ^ { 4 } + 38 y ^ { 3 } + 6 y ^ { 2 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(20y^{4}+19y^{3}+3y^{2}\right)$$

Consider 20y^{4}+19y^{3}+3y^{2}. Factor out y^{2}.

$$y^{2}\left(20y^{2}+19y+3\right)$$

Consider 20y^{2}+19y+3. Factor the expression by grouping. First, the expression needs to be rewritten as 20y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=19 ab=20\times 3=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 19.

$$a=4 b=15$$

Rewrite 20y^{2}+19y+3 as \left(20y^{2}+4y\right)+\left(15y+3\right).

$$\left(20y^{2}+4y\right)+\left(15y+3\right)$$

Factor out 4y in the first and 3 in the second group.

$$4y\left(5y+1\right)+3\left(5y+1\right)$$

Factor out common term 5y+1 by using distributive property.

$$\left(5y+1\right)\left(4y+3\right)$$

Rewrite the complete factored expression.

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

Evaluate

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