Solve 16v-4-15v^2=

$$16 v - 4 - 15 v ^ { 2 } =$$

Factor

$$\left(2-5v\right)\left(3v-2\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-15v^{2}+16v-4$$

Factor the expression by grouping. First, the expression needs to be rewritten as -15v^{2}+av+bv-4. To find a and b, set up a system to be solved.

$$a+b=16 ab=-15\left(-4\right)=60$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=10 b=6$$

Rewrite -15v^{2}+16v-4 as \left(-15v^{2}+10v\right)+\left(6v-4\right).

$$\left(-15v^{2}+10v\right)+\left(6v-4\right)$$

Factor out -5v in the first and 2 in the second group.

$$-5v\left(3v-2\right)+2\left(3v-2\right)$$

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

$$\left(3v-2\right)\left(-5v+2\right)$$

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.

$$-15v^{2}+16v-4=0$$

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.

$$v=\frac{-16±\sqrt{16^{2}-4\left(-15\right)\left(-4\right)}}{2\left(-15\right)}$$

Square 16.

$$v=\frac{-16±\sqrt{256-4\left(-15\right)\left(-4\right)}}{2\left(-15\right)}$$

Multiply -4 times -15.

$$v=\frac{-16±\sqrt{256+60\left(-4\right)}}{2\left(-15\right)}$$

Multiply 60 times -4.

$$v=\frac{-16±\sqrt{256-240}}{2\left(-15\right)}$$

Add 256 to -240.

$$v=\frac{-16±\sqrt{16}}{2\left(-15\right)}$$

Take the square root of 16.

$$v=\frac{-16±4}{2\left(-15\right)}$$

Multiply 2 times -15.

$$v=\frac{-16±4}{-30}$$

Now solve the equation v=\frac{-16±4}{-30} when ± is plus. Add -16 to 4.

$$v=-\frac{12}{-30}$$

Reduce the fraction \frac{-12}{-30} to lowest terms by extracting and canceling out 6.

$$v=\frac{2}{5}$$

Now solve the equation v=\frac{-16±4}{-30} when ± is minus. Subtract 4 from -16.

$$v=-\frac{20}{-30}$$

Reduce the fraction \frac{-20}{-30} to lowest terms by extracting and canceling out 10.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and \frac{2}{3} for x_{2}.

$$-15v^{2}+16v-4=-15\left(v-\frac{2}{5}\right)\left(v-\frac{2}{3}\right)$$

Subtract \frac{2}{5} from v by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-15v^{2}+16v-4=-15\times \frac{-5v+2}{-5}\left(v-\frac{2}{3}\right)$$

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

$$-15v^{2}+16v-4=-15\times \frac{-5v+2}{-5}\times \frac{-3v+2}{-3}$$

Multiply \frac{-5v+2}{-5} times \frac{-3v+2}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-15v^{2}+16v-4=-15\times \frac{\left(-5v+2\right)\left(-3v+2\right)}{-5\left(-3\right)}$$

Multiply -5 times -3.

$$-15v^{2}+16v-4=-15\times \frac{\left(-5v+2\right)\left(-3v+2\right)}{15}$$

Cancel out 15, the greatest common factor in -15 and 15.

$$-15v^{2}+16v-4=-\left(-5v+2\right)\left(-3v+2\right)$$

Evaluate

$$\left(2-5v\right)\left(3v-2\right)$$