Solve 24y-16y^2

$$24y-16 { y }^{ 2 }$$

Factor

$$8y\left(3-2y\right)$$

Solution Steps

Factor out 8.

$$8\left(3y-2y^{2}\right)$$

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

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

Rewrite the complete factored expression.

$$8y\left(-2y+3\right)$$

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.

$$-16y^{2}+24y=0$$

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.

$$y=\frac{-24±\sqrt{24^{2}}}{2\left(-16\right)}$$

Take the square root of 24^{2}.

$$y=\frac{-24±24}{2\left(-16\right)}$$

Multiply 2 times -16.

$$y=\frac{-24±24}{-32}$$

Now solve the equation y=\frac{-24±24}{-32} when ± is plus. Add -24 to 24.

$$y=\frac{0}{-32}$$

Divide 0 by -32.

$$y=0$$

Now solve the equation y=\frac{-24±24}{-32} when ± is minus. Subtract 24 from -24.

$$y=-\frac{48}{-32}$$

Reduce the fraction \frac{-48}{-32} to lowest terms by extracting and canceling out 16.

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

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{3}{2} for x_{2}.

$$-16y^{2}+24y=-16y\left(y-\frac{3}{2}\right)$$

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

$$-16y^{2}+24y=-16y\times \frac{-2y+3}{-2}$$

Cancel out 2, the greatest common factor in -16 and -2.

$$-16y^{2}+24y=8y\left(-2y+3\right)$$

Evaluate

$$8y\left(3-2y\right)$$