Solve y^2-17/4y+1

$$y ^ { 2 } - \frac { 17 } { 4 } y + 1$$

Factor

$$\frac{\left(y-4\right)\left(4y-1\right)}{4}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{4y^{2}-17y+4}{4}$$

Consider 4y^{2}-17y+4. Factor the expression by grouping. First, the expression needs to be rewritten as 4y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=-17 ab=4\times 4=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -17.

$$a=-16 b=-1$$

Rewrite 4y^{2}-17y+4 as \left(4y^{2}-16y\right)+\left(-y+4\right).

$$\left(4y^{2}-16y\right)+\left(-y+4\right)$$

Factor out 4y in the first and -1 in the second group.

$$4y\left(y-4\right)-\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(4y-1\right)$$

Rewrite the complete factored expression.

$$\frac{\left(y-4\right)\left(4y-1\right)}{4}$$

Evaluate

$$y^{2}-\frac{17y}{4}+1$$