Solve 15x^2-63x+54

$$15 x ^ { 2 } - 63 x + 54$$

Factor

$$3\left(x-3\right)\left(5x-6\right)$$

Solution Steps

Factor out 3.

$$3\left(5x^{2}-21x+18\right)$$

Consider 5x^{2}-21x+18. Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+18. To find a and b, set up a system to be solved.

$$a+b=-21 ab=5\times 18=90$$

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 90.

$$-1,-90 -2,-45 -3,-30 -5,-18 -6,-15 -9,-10$$

Calculate the sum for each pair.

$$-1-90=-91 -2-45=-47 -3-30=-33 -5-18=-23 -6-15=-21 -9-10=-19$$

The solution is the pair that gives sum -21.

$$a=-15 b=-6$$

Rewrite 5x^{2}-21x+18 as \left(5x^{2}-15x\right)+\left(-6x+18\right).

$$\left(5x^{2}-15x\right)+\left(-6x+18\right)$$

Factor out 5x in the first and -6 in the second group.

$$5x\left(x-3\right)-6\left(x-3\right)$$

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

$$\left(x-3\right)\left(5x-6\right)$$

Rewrite the complete factored expression.

$$3\left(x-3\right)\left(5x-6\right)$$

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.

$$15x^{2}-63x+54=0$$

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.

$$x=\frac{-\left(-63\right)±\sqrt{\left(-63\right)^{2}-4\times 15\times 54}}{2\times 15}$$

Square -63.

$$x=\frac{-\left(-63\right)±\sqrt{3969-4\times 15\times 54}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-63\right)±\sqrt{3969-60\times 54}}{2\times 15}$$

Multiply -60 times 54.

$$x=\frac{-\left(-63\right)±\sqrt{3969-3240}}{2\times 15}$$

Add 3969 to -3240.

$$x=\frac{-\left(-63\right)±\sqrt{729}}{2\times 15}$$

Take the square root of 729.

$$x=\frac{-\left(-63\right)±27}{2\times 15}$$

The opposite of -63 is 63.

$$x=\frac{63±27}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{63±27}{30}$$

Now solve the equation x=\frac{63±27}{30} when ± is plus. Add 63 to 27.

$$x=\frac{90}{30}$$

Divide 90 by 30.

$$x=3$$

Now solve the equation x=\frac{63±27}{30} when ± is minus. Subtract 27 from 63.

$$x=\frac{36}{30}$$

Reduce the fraction \frac{36}{30} to lowest terms by extracting and canceling out 6.

$$x=\frac{6}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{6}{5} for x_{2}.

$$15x^{2}-63x+54=15\left(x-3\right)\left(x-\frac{6}{5}\right)$$

Subtract \frac{6}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-63x+54=15\left(x-3\right)\times \frac{5x-6}{5}$$

Cancel out 5, the greatest common factor in 15 and 5.

$$15x^{2}-63x+54=3\left(x-3\right)\left(5x-6\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 15

$$x ^ 2 -\frac{21}{5}x +\frac{18}{5} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{21}{5} rs = \frac{18}{5}$$

Two numbers r and s sum up to \frac{21}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{21}{5} = \frac{21}{10}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{21}{10} - u s = \frac{21}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{18}{5}

$$(\frac{21}{10} - u) (\frac{21}{10} + u) = \frac{18}{5}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{441}{100} - u^2 = \frac{18}{5}$$

Simplify the expression by subtracting \frac{441}{100} on both sides

$$-u^2 = \frac{18}{5}-\frac{441}{100} = -\frac{81}{100}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{81}{100} u = \pm\sqrt{\frac{81}{100}} = \pm \frac{9}{10}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{21}{10} - \frac{9}{10} = 1.200 s = \frac{21}{10} + \frac{9}{10} = 3$$

Evaluate

$$3\left(x-3\right)\left(5x-6\right)$$