Solve 45x^2-69x+12

$$45 x ^ { 2 } - 69 x + 12$$

Factor

$$3\left(3x-4\right)\left(5x-1\right)$$

Solution Steps

Factor out 3.

$$3\left(15x^{2}-23x+4\right)$$

Consider 15x^{2}-23x+4. Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-23 ab=15\times 4=60$$

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 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 -23.

$$a=-20 b=-3$$

Rewrite 15x^{2}-23x+4 as \left(15x^{2}-20x\right)+\left(-3x+4\right).

$$\left(15x^{2}-20x\right)+\left(-3x+4\right)$$

Factor out 5x in the first and -1 in the second group.

$$5x\left(3x-4\right)-\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(5x-1\right)$$

Rewrite the complete factored expression.

$$3\left(3x-4\right)\left(5x-1\right)$$

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.

$$45x^{2}-69x+12=0$$

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.

$$x=\frac{-\left(-69\right)±\sqrt{\left(-69\right)^{2}-4\times 45\times 12}}{2\times 45}$$

Square -69.

$$x=\frac{-\left(-69\right)±\sqrt{4761-4\times 45\times 12}}{2\times 45}$$

Multiply -4 times 45.

$$x=\frac{-\left(-69\right)±\sqrt{4761-180\times 12}}{2\times 45}$$

Multiply -180 times 12.

$$x=\frac{-\left(-69\right)±\sqrt{4761-2160}}{2\times 45}$$

Add 4761 to -2160.

$$x=\frac{-\left(-69\right)±\sqrt{2601}}{2\times 45}$$

Take the square root of 2601.

$$x=\frac{-\left(-69\right)±51}{2\times 45}$$

The opposite of -69 is 69.

$$x=\frac{69±51}{2\times 45}$$

Multiply 2 times 45.

$$x=\frac{69±51}{90}$$

Now solve the equation x=\frac{69±51}{90} when ± is plus. Add 69 to 51.

$$x=\frac{120}{90}$$

Reduce the fraction \frac{120}{90} to lowest terms by extracting and canceling out 30.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{69±51}{90} when ± is minus. Subtract 51 from 69.

$$x=\frac{18}{90}$$

Reduce the fraction \frac{18}{90} to lowest terms by extracting and canceling out 18.

$$x=\frac{1}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and \frac{1}{5} for x_{2}.

$$45x^{2}-69x+12=45\left(x-\frac{4}{3}\right)\left(x-\frac{1}{5}\right)$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$45x^{2}-69x+12=45\times \frac{3x-4}{3}\left(x-\frac{1}{5}\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$45x^{2}-69x+12=45\times \frac{3x-4}{3}\times \frac{5x-1}{5}$$

Multiply \frac{3x-4}{3} times \frac{5x-1}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$45x^{2}-69x+12=45\times \frac{\left(3x-4\right)\left(5x-1\right)}{3\times 5}$$

Multiply 3 times 5.

$$45x^{2}-69x+12=45\times \frac{\left(3x-4\right)\left(5x-1\right)}{15}$$

Cancel out 15, the greatest common factor in 45 and 15.

$$45x^{2}-69x+12=3\left(3x-4\right)\left(5x-1\right)$$

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 45

$$x ^ 2 -\frac{23}{15}x +\frac{4}{15} = 0$$

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

$$r + s = \frac{23}{15} rs = \frac{4}{15}$$

Two numbers r and s sum up to \frac{23}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{15} = \frac{23}{30}. 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>

$$r = \frac{23}{30} - u s = \frac{23}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{15}

$$(\frac{23}{30} - u) (\frac{23}{30} + u) = \frac{4}{15}$$

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

$$\frac{529}{900} - u^2 = \frac{4}{15}$$

Simplify the expression by subtracting \frac{529}{900} on both sides

$$-u^2 = \frac{4}{15}-\frac{529}{900} = -\frac{289}{900}$$

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

$$u^2 = \frac{289}{900} u = \pm\sqrt{\frac{289}{900}} = \pm \frac{17}{30}$$

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

$$r =\frac{23}{30} - \frac{17}{30} = 0.200 s = \frac{23}{30} + \frac{17}{30} = 1.333$$

Evaluate

$$45x^{2}-69x+12$$