Solve (3y^2-7y+4)

$$(3 { y }^{ 2 } -7y+4)$$

Factor

$$\left(y-1\right)\left(3y-4\right)$$

Solution Steps

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

$$a+b=-7 ab=3\times 4=12$$

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 -7.

$$a=-4 b=-3$$

Rewrite 3y^{2}-7y+4 as \left(3y^{2}-4y\right)+\left(-3y+4\right).

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

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

$$y\left(3y-4\right)-\left(3y-4\right)$$

Factor out common term 3y-4 by using distributive property.

$$\left(3y-4\right)\left(y-1\right)$$

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.

$$3y^{2}-7y+4=0$$

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.

$$y=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Square -7.

$$y=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-7\right)±\sqrt{49-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$y=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 3}$$

Add 49 to -48.

$$y=\frac{-\left(-7\right)±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$y=\frac{-\left(-7\right)±1}{2\times 3}$$

The opposite of -7 is 7.

$$y=\frac{7±1}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{7±1}{6}$$

Now solve the equation y=\frac{7±1}{6} when ± is plus. Add 7 to 1.

$$y=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$y=\frac{4}{3}$$

Now solve the equation y=\frac{7±1}{6} when ± is minus. Subtract 1 from 7.

$$y=\frac{6}{6}$$

Divide 6 by 6.

$$y=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and 1 for x_{2}.

$$3y^{2}-7y+4=3\left(y-\frac{4}{3}\right)\left(y-1\right)$$

Subtract \frac{4}{3} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3y^{2}-7y+4=3\times \frac{3y-4}{3}\left(y-1\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3y^{2}-7y+4=\left(3y-4\right)\left(y-1\right)$$

Evaluate

$$\left(y-1\right)\left(3y-4\right)$$