Solve 16y^2-48y-160-0=

$$16 { y }^{ 2 } -48y-160-0=$$

Evaluate

$$16\left(y-5\right)\left(y+2\right)$$

Solution Steps

Multiply -1 and 0 to get 0.

$$16y^{2}-48y-160+0$$

Add -160 and 0 to get -160.

$$16y^{2}-48y-160$$

Factor

$$16\left(y-5\right)\left(y+2\right)$$

Solution Steps

Factor out 16.

$$16\left(y^{2}-3y-10\right)$$

Consider y^{2}-3y-10. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-10\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite y^{2}-3y-10 as \left(y^{2}-5y\right)+\left(2y-10\right).

$$\left(y^{2}-5y\right)+\left(2y-10\right)$$

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

$$y\left(y-5\right)+2\left(y-5\right)$$

Factor out common term y-5 by using distributive property.

$$\left(y-5\right)\left(y+2\right)$$

Rewrite the complete factored expression.

$$16\left(y-5\right)\left(y+2\right)$$

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.

$$16y^{2}-48y-160=0$$

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.

$$y=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 16\left(-160\right)}}{2\times 16}$$

Square -48.

$$y=\frac{-\left(-48\right)±\sqrt{2304-4\times 16\left(-160\right)}}{2\times 16}$$

Multiply -4 times 16.

$$y=\frac{-\left(-48\right)±\sqrt{2304-64\left(-160\right)}}{2\times 16}$$

Multiply -64 times -160.

$$y=\frac{-\left(-48\right)±\sqrt{2304+10240}}{2\times 16}$$

Add 2304 to 10240.

$$y=\frac{-\left(-48\right)±\sqrt{12544}}{2\times 16}$$

Take the square root of 12544.

$$y=\frac{-\left(-48\right)±112}{2\times 16}$$

The opposite of -48 is 48.

$$y=\frac{48±112}{2\times 16}$$

Multiply 2 times 16.

$$y=\frac{48±112}{32}$$

Now solve the equation y=\frac{48±112}{32} when ± is plus. Add 48 to 112.

$$y=\frac{160}{32}$$

Divide 160 by 32.

$$y=5$$

Now solve the equation y=\frac{48±112}{32} when ± is minus. Subtract 112 from 48.

$$y=-\frac{64}{32}$$

Divide -64 by 32.

$$y=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -2 for x_{2}.

$$16y^{2}-48y-160=16\left(y-5\right)\left(y-\left(-2\right)\right)$$

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

$$16y^{2}-48y-160=16\left(y-5\right)\left(y+2\right)$$