Solve 4x^2-16x+15

$$4 x ^ { 2 } - 16 x + 15$$

Factor

$$\left(2x-5\right)\left(2x-3\right)$$

Solution Steps

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

$$a+b=-16 ab=4\times 15=60$$

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 -16.

$$a=-10 b=-6$$

Rewrite 4x^{2}-16x+15 as \left(4x^{2}-10x\right)+\left(-6x+15\right).

$$\left(4x^{2}-10x\right)+\left(-6x+15\right)$$

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

$$2x\left(2x-5\right)-3\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(2x-3\right)$$

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.

$$4x^{2}-16x+15=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 15}}{2\times 4}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\times 15}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-16\right)±\sqrt{256-16\times 15}}{2\times 4}$$

Multiply -16 times 15.

$$x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 4}$$

Add 256 to -240.

$$x=\frac{-\left(-16\right)±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$x=\frac{-\left(-16\right)±4}{2\times 4}$$

The opposite of -16 is 16.

$$x=\frac{16±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{16±4}{8}$$

Now solve the equation x=\frac{16±4}{8} when ± is plus. Add 16 to 4.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{16±4}{8} when ± is minus. Subtract 4 from 16.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

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

$$4x^{2}-16x+15=4\left(x-\frac{5}{2}\right)\left(x-\frac{3}{2}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-16x+15=4\times \frac{2x-5}{2}\left(x-\frac{3}{2}\right)$$

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

$$4x^{2}-16x+15=4\times \frac{2x-5}{2}\times \frac{2x-3}{2}$$

Multiply \frac{2x-5}{2} times \frac{2x-3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-16x+15=4\times \frac{\left(2x-5\right)\left(2x-3\right)}{2\times 2}$$

Multiply 2 times 2.

$$4x^{2}-16x+15=4\times \frac{\left(2x-5\right)\left(2x-3\right)}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4x^{2}-16x+15=\left(2x-5\right)\left(2x-3\right)$$

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 4

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

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

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

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. 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>

$$r = 2 - u s = 2 + u$$

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

$$(2 - u) (2 + u) = \frac{15}{4}$$

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

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

Simplify the expression by subtracting 4 on both sides

$$-u^2 = \frac{15}{4}-4 = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =2 - \frac{1}{2} = 1.500 s = 2 + \frac{1}{2} = 2.500$$

Evaluate

$$\left(2x-5\right)\left(2x-3\right)$$