Solve 25a^2-35a+12

$$25 a ^ { 2 } - 35 a + 12$$

Factor

$$\left(5a-4\right)\left(5a-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25a^{2}+pa+qa+12. To find p and q, set up a system to be solved.

$$p+q=-35 pq=25\times 12=300$$

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 300.

$$-1,-300 -2,-150 -3,-100 -4,-75 -5,-60 -6,-50 -10,-30 -12,-25 -15,-20$$

Calculate the sum for each pair.

$$-1-300=-301 -2-150=-152 -3-100=-103 -4-75=-79 -5-60=-65 -6-50=-56 -10-30=-40 -12-25=-37 -15-20=-35$$

The solution is the pair that gives sum -35.

$$p=-20 q=-15$$

Rewrite 25a^{2}-35a+12 as \left(25a^{2}-20a\right)+\left(-15a+12\right).

$$\left(25a^{2}-20a\right)+\left(-15a+12\right)$$

Factor out 5a in the first and -3 in the second group.

$$5a\left(5a-4\right)-3\left(5a-4\right)$$

Factor out common term 5a-4 by using distributive property.

$$\left(5a-4\right)\left(5a-3\right)$$

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.

$$25a^{2}-35a+12=0$$

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.

$$a=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 25\times 12}}{2\times 25}$$

Square -35.

$$a=\frac{-\left(-35\right)±\sqrt{1225-4\times 25\times 12}}{2\times 25}$$

Multiply -4 times 25.

$$a=\frac{-\left(-35\right)±\sqrt{1225-100\times 12}}{2\times 25}$$

Multiply -100 times 12.

$$a=\frac{-\left(-35\right)±\sqrt{1225-1200}}{2\times 25}$$

Add 1225 to -1200.

$$a=\frac{-\left(-35\right)±\sqrt{25}}{2\times 25}$$

Take the square root of 25.

$$a=\frac{-\left(-35\right)±5}{2\times 25}$$

The opposite of -35 is 35.

$$a=\frac{35±5}{2\times 25}$$

Multiply 2 times 25.

$$a=\frac{35±5}{50}$$

Now solve the equation a=\frac{35±5}{50} when ± is plus. Add 35 to 5.

$$a=\frac{40}{50}$$

Reduce the fraction \frac{40}{50} to lowest terms by extracting and canceling out 10.

$$a=\frac{4}{5}$$

Now solve the equation a=\frac{35±5}{50} when ± is minus. Subtract 5 from 35.

$$a=\frac{30}{50}$$

Reduce the fraction \frac{30}{50} to lowest terms by extracting and canceling out 10.

$$a=\frac{3}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{5} for x_{1} and \frac{3}{5} for x_{2}.

$$25a^{2}-35a+12=25\left(a-\frac{4}{5}\right)\left(a-\frac{3}{5}\right)$$

Subtract \frac{4}{5} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$25a^{2}-35a+12=25\times \frac{5a-4}{5}\left(a-\frac{3}{5}\right)$$

Subtract \frac{3}{5} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$25a^{2}-35a+12=25\times \frac{5a-4}{5}\times \frac{5a-3}{5}$$

Multiply \frac{5a-4}{5} times \frac{5a-3}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25a^{2}-35a+12=25\times \frac{\left(5a-4\right)\left(5a-3\right)}{5\times 5}$$

Multiply 5 times 5.

$$25a^{2}-35a+12=25\times \frac{\left(5a-4\right)\left(5a-3\right)}{25}$$

Cancel out 25, the greatest common factor in 25 and 25.

$$25a^{2}-35a+12=\left(5a-4\right)\left(5a-3\right)$$

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 25

$$x ^ 2 -\frac{7}{5}x +\frac{12}{25} = 0$$

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

$$r + s = \frac{7}{5} rs = \frac{12}{25}$$

Two numbers r and s sum up to \frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{5} = \frac{7}{10}. 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>

$$r = \frac{7}{10} - u s = \frac{7}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{12}{25}

$$(\frac{7}{10} - u) (\frac{7}{10} + u) = \frac{12}{25}$$

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

$$\frac{49}{100} - u^2 = \frac{12}{25}$$

Simplify the expression by subtracting \frac{49}{100} on both sides

$$-u^2 = \frac{12}{25}-\frac{49}{100} = -\frac{1}{100}$$

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

$$u^2 = \frac{1}{100} u = \pm\sqrt{\frac{1}{100}} = \pm \frac{1}{10}$$

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

$$r =\frac{7}{10} - \frac{1}{10} = 0.600 s = \frac{7}{10} + \frac{1}{10} = 0.800$$

Evaluate

$$\left(5a-4\right)\left(5a-3\right)$$