Solve 15x^2-23x+8

$$15 x ^ { 2 } - 23 x + 8$$

Factor

$$\left(x-1\right)\left(15x-8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-23 ab=15\times 8=120$$

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 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -23.

$$a=-15 b=-8$$

Rewrite 15x^{2}-23x+8 as \left(15x^{2}-15x\right)+\left(-8x+8\right).

$$\left(15x^{2}-15x\right)+\left(-8x+8\right)$$

Factor out 15x in the first and -8 in the second group.

$$15x\left(x-1\right)-8\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(15x-8\right)$$

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.

$$15x^{2}-23x+8=0$$

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.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 15\times 8}}{2\times 15}$$

Square -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 15\times 8}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-23\right)±\sqrt{529-60\times 8}}{2\times 15}$$

Multiply -60 times 8.

$$x=\frac{-\left(-23\right)±\sqrt{529-480}}{2\times 15}$$

Add 529 to -480.

$$x=\frac{-\left(-23\right)±\sqrt{49}}{2\times 15}$$

Take the square root of 49.

$$x=\frac{-\left(-23\right)±7}{2\times 15}$$

The opposite of -23 is 23.

$$x=\frac{23±7}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{23±7}{30}$$

Now solve the equation x=\frac{23±7}{30} when ± is plus. Add 23 to 7.

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

Divide 30 by 30.

$$x=1$$

Now solve the equation x=\frac{23±7}{30} when ± is minus. Subtract 7 from 23.

$$x=\frac{16}{30}$$

Reduce the fraction \frac{16}{30} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{15}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and \frac{8}{15} for x_{2}.

$$15x^{2}-23x+8=15\left(x-1\right)\left(x-\frac{8}{15}\right)$$

Subtract \frac{8}{15} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-23x+8=15\left(x-1\right)\times \frac{15x-8}{15}$$

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

$$15x^{2}-23x+8=\left(x-1\right)\left(15x-8\right)$$

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

$$x ^ 2 -\frac{23}{15}x +\frac{8}{15} = 0$$

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

$$r + s = \frac{23}{15} rs = \frac{8}{15}$$

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{8}{15}

$$(\frac{23}{30} - u) (\frac{23}{30} + u) = \frac{8}{15}$$

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

$$\frac{529}{900} - u^2 = \frac{8}{15}$$

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

$$-u^2 = \frac{8}{15}-\frac{529}{900} = -\frac{49}{900}$$

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

$$u^2 = \frac{49}{900} u = \pm\sqrt{\frac{49}{900}} = \pm \frac{7}{30}$$

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

$$r =\frac{23}{30} - \frac{7}{30} = 0.533 s = \frac{23}{30} + \frac{7}{30} = 1.000$$

Evaluate

$$\left(x-1\right)\left(15x-8\right)$$