Solve =6a^2-25a+24

$$= 6 a ^ { 2 } - 25 a + 24$$

Factor

$$\left(2a-3\right)\left(3a-8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa+24. To find p and q, set up a system to be solved.

$$p+q=-25 pq=6\times 24=144$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 144.

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

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -25.

$$p=-16 q=-9$$

Rewrite 6a^{2}-25a+24 as \left(6a^{2}-16a\right)+\left(-9a+24\right).

$$\left(6a^{2}-16a\right)+\left(-9a+24\right)$$

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

$$2a\left(3a-8\right)-3\left(3a-8\right)$$

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

$$\left(3a-8\right)\left(2a-3\right)$$

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.

$$6a^{2}-25a+24=0$$

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.

$$a=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 6\times 24}}{2\times 6}$$

Square -25.

$$a=\frac{-\left(-25\right)±\sqrt{625-4\times 6\times 24}}{2\times 6}$$

Multiply -4 times 6.

$$a=\frac{-\left(-25\right)±\sqrt{625-24\times 24}}{2\times 6}$$

Multiply -24 times 24.

$$a=\frac{-\left(-25\right)±\sqrt{625-576}}{2\times 6}$$

Add 625 to -576.

$$a=\frac{-\left(-25\right)±\sqrt{49}}{2\times 6}$$

Take the square root of 49.

$$a=\frac{-\left(-25\right)±7}{2\times 6}$$

The opposite of -25 is 25.

$$a=\frac{25±7}{2\times 6}$$

Multiply 2 times 6.

$$a=\frac{25±7}{12}$$

Now solve the equation a=\frac{25±7}{12} when ± is plus. Add 25 to 7.

$$a=\frac{32}{12}$$

Reduce the fraction \frac{32}{12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation a=\frac{25±7}{12} when ± is minus. Subtract 7 from 25.

$$a=\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

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

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

$$6a^{2}-25a+24=6\left(a-\frac{8}{3}\right)\left(a-\frac{3}{2}\right)$$

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

$$6a^{2}-25a+24=6\times \frac{3a-8}{3}\left(a-\frac{3}{2}\right)$$

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

$$6a^{2}-25a+24=6\times \frac{3a-8}{3}\times \frac{2a-3}{2}$$

Multiply \frac{3a-8}{3} times \frac{2a-3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6a^{2}-25a+24=6\times \frac{\left(3a-8\right)\left(2a-3\right)}{3\times 2}$$

Multiply 3 times 2.

$$6a^{2}-25a+24=6\times \frac{\left(3a-8\right)\left(2a-3\right)}{6}$$

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

$$6a^{2}-25a+24=\left(3a-8\right)\left(2a-3\right)$$

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 6

$$x ^ 2 -\frac{25}{6}x +4 = 0$$

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

$$r + s = \frac{25}{6} rs = 4$$

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 = 4

$$(\frac{25}{12} - u) (\frac{25}{12} + u) = 4$$

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

$$\frac{625}{144} - u^2 = 4$$

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

$$-u^2 = 4-\frac{625}{144} = -\frac{49}{144}$$

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

$$u^2 = \frac{49}{144} u = \pm\sqrt{\frac{49}{144}} = \pm \frac{7}{12}$$

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

$$r =\frac{25}{12} - \frac{7}{12} = 1.500 s = \frac{25}{12} + \frac{7}{12} = 2.667$$

Evaluate

$$\left(2a-3\right)\left(3a-8\right)$$