Solve 4y^2-8y+3

$$4 { y }^{ 2 } -8y+3$$

Factor

$$\left(2y-3\right)\left(2y-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=-8 ab=4\times 3=12$$

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 12.

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

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite 4y^{2}-8y+3 as \left(4y^{2}-6y\right)+\left(-2y+3\right).

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

Factor out 2y in the first and -1 in the second group.

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

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(2y-1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4y^{2}-8y+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-\left(-8\right)±\sqrt{64-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$y=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 4}$$

Add 64 to -48.

$$y=\frac{-\left(-8\right)±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$y=\frac{-\left(-8\right)±4}{2\times 4}$$

The opposite of -8 is 8.

$$y=\frac{8±4}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{8±4}{8}$$

Now solve the equation y=\frac{8±4}{8} when ± is plus. Add 8 to 4.

$$y=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$y=\frac{3}{2}$$

Now solve the equation y=\frac{8±4}{8} when ± is minus. Subtract 4 from 8.

$$y=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

$$y=\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and \frac{1}{2} for x_{2}.

$$4y^{2}-8y+3=4\left(y-\frac{3}{2}\right)\left(y-\frac{1}{2}\right)$$

Subtract \frac{3}{2} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4y^{2}-8y+3=4\times \frac{2y-3}{2}\left(y-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4y^{2}-8y+3=4\times \frac{2y-3}{2}\times \frac{2y-1}{2}$$

Multiply \frac{2y-3}{2} times \frac{2y-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4y^{2}-8y+3=4\times \frac{\left(2y-3\right)\left(2y-1\right)}{2\times 2}$$

Multiply 2 times 2.

$$4y^{2}-8y+3=4\times \frac{\left(2y-3\right)\left(2y-1\right)}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4y^{2}-8y+3=\left(2y-3\right)\left(2y-1\right)$$

Evaluate

$$\left(2y-3\right)\left(2y-1\right)$$