Solve 54x^2-141x+90

$$54 { x }^{ 2 } -141x+90$$

Factor

$$3\left(2x-3\right)\left(9x-10\right)$$

Solution Steps

Factor out 3.

$$3\left(18x^{2}-47x+30\right)$$

Consider 18x^{2}-47x+30. Factor the expression by grouping. First, the expression needs to be rewritten as 18x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=-47 ab=18\times 30=540$$

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 540.

$$-1,-540 -2,-270 -3,-180 -4,-135 -5,-108 -6,-90 -9,-60 -10,-54 -12,-45 -15,-36 -18,-30 -20,-27$$

Calculate the sum for each pair.

$$-1-540=-541 -2-270=-272 -3-180=-183 -4-135=-139 -5-108=-113 -6-90=-96 -9-60=-69 -10-54=-64 -12-45=-57 -15-36=-51 -18-30=-48 -20-27=-47$$

The solution is the pair that gives sum -47.

$$a=-27 b=-20$$

Rewrite 18x^{2}-47x+30 as \left(18x^{2}-27x\right)+\left(-20x+30\right).

$$\left(18x^{2}-27x\right)+\left(-20x+30\right)$$

Factor out 9x in the first and -10 in the second group.

$$9x\left(2x-3\right)-10\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(9x-10\right)$$

Rewrite the complete factored expression.

$$3\left(2x-3\right)\left(9x-10\right)$$

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.

$$54x^{2}-141x+90=0$$

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.

$$x=\frac{-\left(-141\right)±\sqrt{\left(-141\right)^{2}-4\times 54\times 90}}{2\times 54}$$

Square -141.

$$x=\frac{-\left(-141\right)±\sqrt{19881-4\times 54\times 90}}{2\times 54}$$

Multiply -4 times 54.

$$x=\frac{-\left(-141\right)±\sqrt{19881-216\times 90}}{2\times 54}$$

Multiply -216 times 90.

$$x=\frac{-\left(-141\right)±\sqrt{19881-19440}}{2\times 54}$$

Add 19881 to -19440.

$$x=\frac{-\left(-141\right)±\sqrt{441}}{2\times 54}$$

Take the square root of 441.

$$x=\frac{-\left(-141\right)±21}{2\times 54}$$

The opposite of -141 is 141.

$$x=\frac{141±21}{2\times 54}$$

Multiply 2 times 54.

$$x=\frac{141±21}{108}$$

Now solve the equation x=\frac{141±21}{108} when ± is plus. Add 141 to 21.

$$x=\frac{162}{108}$$

Reduce the fraction \frac{162}{108} to lowest terms by extracting and canceling out 54.

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

Now solve the equation x=\frac{141±21}{108} when ± is minus. Subtract 21 from 141.

$$x=\frac{120}{108}$$

Reduce the fraction \frac{120}{108} to lowest terms by extracting and canceling out 12.

$$x=\frac{10}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and \frac{10}{9} for x_{2}.

$$54x^{2}-141x+90=54\left(x-\frac{3}{2}\right)\left(x-\frac{10}{9}\right)$$

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

$$54x^{2}-141x+90=54\times \frac{2x-3}{2}\left(x-\frac{10}{9}\right)$$

Subtract \frac{10}{9} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$54x^{2}-141x+90=54\times \frac{2x-3}{2}\times \frac{9x-10}{9}$$

Multiply \frac{2x-3}{2} times \frac{9x-10}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$54x^{2}-141x+90=54\times \frac{\left(2x-3\right)\left(9x-10\right)}{2\times 9}$$

Multiply 2 times 9.

$$54x^{2}-141x+90=54\times \frac{\left(2x-3\right)\left(9x-10\right)}{18}$$

Cancel out 18, the greatest common factor in 54 and 18.

$$54x^{2}-141x+90=3\left(2x-3\right)\left(9x-10\right)$$

Evaluate

$$54x^{2}-141x+90$$