Solve 12x^2-41x+24

$$12 x ^ { 2 } - 41 x + 24$$

Factor

$$\left(3x-8\right)\left(4x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-41 ab=12\times 24=288$$

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 288.

$$-1,-288 -2,-144 -3,-96 -4,-72 -6,-48 -8,-36 -9,-32 -12,-24 -16,-18$$

Calculate the sum for each pair.

$$-1-288=-289 -2-144=-146 -3-96=-99 -4-72=-76 -6-48=-54 -8-36=-44 -9-32=-41 -12-24=-36 -16-18=-34$$

The solution is the pair that gives sum -41.

$$a=-32 b=-9$$

Rewrite 12x^{2}-41x+24 as \left(12x^{2}-32x\right)+\left(-9x+24\right).

$$\left(12x^{2}-32x\right)+\left(-9x+24\right)$$

Factor out 4x in the first and -3 in the second group.

$$4x\left(3x-8\right)-3\left(3x-8\right)$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(4x-3\right)$$

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.

$$12x^{2}-41x+24=0$$

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.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 12\times 24}}{2\times 12}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 12\times 24}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-41\right)±\sqrt{1681-48\times 24}}{2\times 12}$$

Multiply -48 times 24.

$$x=\frac{-\left(-41\right)±\sqrt{1681-1152}}{2\times 12}$$

Add 1681 to -1152.

$$x=\frac{-\left(-41\right)±\sqrt{529}}{2\times 12}$$

Take the square root of 529.

$$x=\frac{-\left(-41\right)±23}{2\times 12}$$

The opposite of -41 is 41.

$$x=\frac{41±23}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{41±23}{24}$$

Now solve the equation x=\frac{41±23}{24} when ± is plus. Add 41 to 23.

$$x=\frac{64}{24}$$

Reduce the fraction \frac{64}{24} to lowest terms by extracting and canceling out 8.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{41±23}{24} when ± is minus. Subtract 23 from 41.

$$x=\frac{18}{24}$$

Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and \frac{3}{4} for x_{2}.

$$12x^{2}-41x+24=12\left(x-\frac{8}{3}\right)\left(x-\frac{3}{4}\right)$$

Subtract \frac{8}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-41x+24=12\times \frac{3x-8}{3}\left(x-\frac{3}{4}\right)$$

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

$$12x^{2}-41x+24=12\times \frac{3x-8}{3}\times \frac{4x-3}{4}$$

Multiply \frac{3x-8}{3} times \frac{4x-3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-41x+24=12\times \frac{\left(3x-8\right)\left(4x-3\right)}{3\times 4}$$

Multiply 3 times 4.

$$12x^{2}-41x+24=12\times \frac{\left(3x-8\right)\left(4x-3\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12x^{2}-41x+24=\left(3x-8\right)\left(4x-3\right)$$

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 12

$$x ^ 2 -\frac{41}{12}x +2 = 0$$

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

$$r + s = \frac{41}{12} rs = 2$$

Two numbers r and s sum up to \frac{41}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{41}{12} = \frac{41}{24}. 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>

$$r = \frac{41}{24} - u s = \frac{41}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 2

$$(\frac{41}{24} - u) (\frac{41}{24} + u) = 2$$

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

$$\frac{1681}{576} - u^2 = 2$$

Simplify the expression by subtracting \frac{1681}{576} on both sides

$$-u^2 = 2-\frac{1681}{576} = -\frac{529}{576}$$

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

$$u^2 = \frac{529}{576} u = \pm\sqrt{\frac{529}{576}} = \pm \frac{23}{24}$$

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

$$r =\frac{41}{24} - \frac{23}{24} = 0.750 s = \frac{41}{24} + \frac{23}{24} = 2.667$$

Evaluate

$$\left(3x-8\right)\left(4x-3\right)$$