Solve 10y^4-26y^3+12y^2

$$10 y ^ { 4 } - 26 y ^ { 3 } + 12 y ^ { 2 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(5y^{4}-13y^{3}+6y^{2}\right)$$

Consider 5y^{4}-13y^{3}+6y^{2}. Factor out y^{2}.

$$y^{2}\left(5y^{2}-13y+6\right)$$

Consider 5y^{2}-13y+6. Factor the expression by grouping. First, the expression needs to be rewritten as 5y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=-13 ab=5\times 6=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

Rewrite 5y^{2}-13y+6 as \left(5y^{2}-10y\right)+\left(-3y+6\right).

$$\left(5y^{2}-10y\right)+\left(-3y+6\right)$$

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

$$5y\left(y-2\right)-3\left(y-2\right)$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(5y-3\right)$$

Rewrite the complete factored expression.

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

Evaluate

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