Solve 9x^2-81x+50

$$9 { x }^{ 2 } -81x+50$$

Factor

$$\left(3x-25\right)\left(3x-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx+50. To find a and b, set up a system to be solved.

$$a+b=-81 ab=9\times 50=450$$

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 450.

$$-1,-450 -2,-225 -3,-150 -5,-90 -6,-75 -9,-50 -10,-45 -15,-30 -18,-25$$

Calculate the sum for each pair.

$$-1-450=-451 -2-225=-227 -3-150=-153 -5-90=-95 -6-75=-81 -9-50=-59 -10-45=-55 -15-30=-45 -18-25=-43$$

The solution is the pair that gives sum -81.

$$a=-75 b=-6$$

Rewrite 9x^{2}-81x+50 as \left(9x^{2}-75x\right)+\left(-6x+50\right).

$$\left(9x^{2}-75x\right)+\left(-6x+50\right)$$

Factor out 3x in the first and -2 in the second group.

$$3x\left(3x-25\right)-2\left(3x-25\right)$$

Factor out common term 3x-25 by using distributive property.

$$\left(3x-25\right)\left(3x-2\right)$$

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.

$$9x^{2}-81x+50=0$$

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.

$$x=\frac{-\left(-81\right)±\sqrt{\left(-81\right)^{2}-4\times 9\times 50}}{2\times 9}$$

Square -81.

$$x=\frac{-\left(-81\right)±\sqrt{6561-4\times 9\times 50}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-81\right)±\sqrt{6561-36\times 50}}{2\times 9}$$

Multiply -36 times 50.

$$x=\frac{-\left(-81\right)±\sqrt{6561-1800}}{2\times 9}$$

Add 6561 to -1800.

$$x=\frac{-\left(-81\right)±\sqrt{4761}}{2\times 9}$$

Take the square root of 4761.

$$x=\frac{-\left(-81\right)±69}{2\times 9}$$

The opposite of -81 is 81.

$$x=\frac{81±69}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{81±69}{18}$$

Now solve the equation x=\frac{81±69}{18} when ± is plus. Add 81 to 69.

$$x=\frac{150}{18}$$

Reduce the fraction \frac{150}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{25}{3}$$

Now solve the equation x=\frac{81±69}{18} when ± is minus. Subtract 69 from 81.

$$x=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{25}{3} for x_{1} and \frac{2}{3} for x_{2}.

$$9x^{2}-81x+50=9\left(x-\frac{25}{3}\right)\left(x-\frac{2}{3}\right)$$

Subtract \frac{25}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-81x+50=9\times \frac{3x-25}{3}\left(x-\frac{2}{3}\right)$$

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

$$9x^{2}-81x+50=9\times \frac{3x-25}{3}\times \frac{3x-2}{3}$$

Multiply \frac{3x-25}{3} times \frac{3x-2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-81x+50=9\times \frac{\left(3x-25\right)\left(3x-2\right)}{3\times 3}$$

Multiply 3 times 3.

$$9x^{2}-81x+50=9\times \frac{\left(3x-25\right)\left(3x-2\right)}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9x^{2}-81x+50=\left(3x-25\right)\left(3x-2\right)$$

Evaluate

$$\left(3x-25\right)\left(3x-2\right)$$