Solve 24x^2-100x+16

$$24 x ^ { 2 } - 100 x + 16$$

Factor

$$4\left(x-4\right)\left(6x-1\right)$$

Solution Steps

Factor out 4.

$$4\left(6x^{2}-25x+4\right)$$

Consider 6x^{2}-25x+4. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-25 ab=6\times 4=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -25.

$$a=-24 b=-1$$

Rewrite 6x^{2}-25x+4 as \left(6x^{2}-24x\right)+\left(-x+4\right).

$$\left(6x^{2}-24x\right)+\left(-x+4\right)$$

Factor out 6x in the first and -1 in the second group.

$$6x\left(x-4\right)-\left(x-4\right)$$

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

$$\left(x-4\right)\left(6x-1\right)$$

Rewrite the complete factored expression.

$$4\left(x-4\right)\left(6x-1\right)$$

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.

$$24x^{2}-100x+16=0$$

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.

$$x=\frac{-\left(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 24\times 16}}{2\times 24}$$

Square -100.

$$x=\frac{-\left(-100\right)±\sqrt{10000-4\times 24\times 16}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-\left(-100\right)±\sqrt{10000-96\times 16}}{2\times 24}$$

Multiply -96 times 16.

$$x=\frac{-\left(-100\right)±\sqrt{10000-1536}}{2\times 24}$$

Add 10000 to -1536.

$$x=\frac{-\left(-100\right)±\sqrt{8464}}{2\times 24}$$

Take the square root of 8464.

$$x=\frac{-\left(-100\right)±92}{2\times 24}$$

The opposite of -100 is 100.

$$x=\frac{100±92}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{100±92}{48}$$

Now solve the equation x=\frac{100±92}{48} when ± is plus. Add 100 to 92.

$$x=\frac{192}{48}$$

Divide 192 by 48.

$$x=4$$

Now solve the equation x=\frac{100±92}{48} when ± is minus. Subtract 92 from 100.

$$x=\frac{8}{48}$$

Reduce the fraction \frac{8}{48} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and \frac{1}{6} for x_{2}.

$$24x^{2}-100x+16=24\left(x-4\right)\left(x-\frac{1}{6}\right)$$

Subtract \frac{1}{6} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}-100x+16=24\left(x-4\right)\times \frac{6x-1}{6}$$

Cancel out 6, the greatest common factor in 24 and 6.

$$24x^{2}-100x+16=4\left(x-4\right)\left(6x-1\right)$$

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 24

$$x ^ 2 -\frac{25}{6}x +\frac{2}{3} = 0$$

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

$$r + s = \frac{25}{6} rs = \frac{2}{3}$$

Two numbers r and s sum up to \frac{25}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{6} = \frac{25}{12}. 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>

$$r = \frac{25}{12} - u s = \frac{25}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}

$$(\frac{25}{12} - u) (\frac{25}{12} + u) = \frac{2}{3}$$

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

$$\frac{625}{144} - u^2 = \frac{2}{3}$$

Simplify the expression by subtracting \frac{625}{144} on both sides

$$-u^2 = \frac{2}{3}-\frac{625}{144} = -\frac{529}{144}$$

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

$$u^2 = \frac{529}{144} u = \pm\sqrt{\frac{529}{144}} = \pm \frac{23}{12}$$

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

$$r =\frac{25}{12} - \frac{23}{12} = 0.167 s = \frac{25}{12} + \frac{23}{12} = 4$$

Evaluate

$$4\left(x-4\right)\left(6x-1\right)$$