Solve 4y+12y^2

$$4 y + 12 y ^ { 2 }$$

Factor

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

Solution Steps

Factor out 4.

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

Consider y+3y^{2}. Factor out y.

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

Rewrite the complete factored expression.

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

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.

$$12y^{2}+4y=0$$

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.

$$y=\frac{-4±\sqrt{4^{2}}}{2\times 12}$$

Take the square root of 4^{2}.

$$y=\frac{-4±4}{2\times 12}$$

Multiply 2 times 12.

$$y=\frac{-4±4}{24}$$

Now solve the equation y=\frac{-4±4}{24} when ± is plus. Add -4 to 4.

$$y=\frac{0}{24}$$

Divide 0 by 24.

$$y=0$$

Now solve the equation y=\frac{-4±4}{24} when ± is minus. Subtract 4 from -4.

$$y=-\frac{8}{24}$$

Reduce the fraction \frac{-8}{24} to lowest terms by extracting and canceling out 8.

$$y=-\frac{1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and -\frac{1}{3} for x_{2}.

$$12y^{2}+4y=12y\left(y-\left(-\frac{1}{3}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$12y^{2}+4y=12y\left(y+\frac{1}{3}\right)$$

Add \frac{1}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12y^{2}+4y=12y\times \frac{3y+1}{3}$$

Cancel out 3, the greatest common factor in 12 and 3.

$$12y^{2}+4y=4y\left(3y+1\right)$$

Evaluate

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