Solve 35x^2+16x-12

$$35 x ^ { 2 } + 16 x - 12$$

Factor

$$\left(5x-2\right)\left(7x+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 35x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=16 ab=35\left(-12\right)=-420$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -420.

$$-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21$$

Calculate the sum for each pair.

$$-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1$$

The solution is the pair that gives sum 16.

$$a=-14 b=30$$

Rewrite 35x^{2}+16x-12 as \left(35x^{2}-14x\right)+\left(30x-12\right).

$$\left(35x^{2}-14x\right)+\left(30x-12\right)$$

Factor out 7x in the first and 6 in the second group.

$$7x\left(5x-2\right)+6\left(5x-2\right)$$

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

$$\left(5x-2\right)\left(7x+6\right)$$

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.

$$35x^{2}+16x-12=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 35\left(-12\right)}}{2\times 35}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 35\left(-12\right)}}{2\times 35}$$

Multiply -4 times 35.

$$x=\frac{-16±\sqrt{256-140\left(-12\right)}}{2\times 35}$$

Multiply -140 times -12.

$$x=\frac{-16±\sqrt{256+1680}}{2\times 35}$$

Add 256 to 1680.

$$x=\frac{-16±\sqrt{1936}}{2\times 35}$$

Take the square root of 1936.

$$x=\frac{-16±44}{2\times 35}$$

Multiply 2 times 35.

$$x=\frac{-16±44}{70}$$

Now solve the equation x=\frac{-16±44}{70} when ± is plus. Add -16 to 44.

$$x=\frac{28}{70}$$

Reduce the fraction \frac{28}{70} to lowest terms by extracting and canceling out 14.

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

Now solve the equation x=\frac{-16±44}{70} when ± is minus. Subtract 44 from -16.

$$x=-\frac{60}{70}$$

Reduce the fraction \frac{-60}{70} to lowest terms by extracting and canceling out 10.

$$x=-\frac{6}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and -\frac{6}{7} for x_{2}.

$$35x^{2}+16x-12=35\left(x-\frac{2}{5}\right)\left(x-\left(-\frac{6}{7}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$35x^{2}+16x-12=35\left(x-\frac{2}{5}\right)\left(x+\frac{6}{7}\right)$$

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

$$35x^{2}+16x-12=35\times \frac{5x-2}{5}\left(x+\frac{6}{7}\right)$$

Add \frac{6}{7} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$35x^{2}+16x-12=35\times \frac{5x-2}{5}\times \frac{7x+6}{7}$$

Multiply \frac{5x-2}{5} times \frac{7x+6}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$35x^{2}+16x-12=35\times \frac{\left(5x-2\right)\left(7x+6\right)}{5\times 7}$$

Multiply 5 times 7.

$$35x^{2}+16x-12=35\times \frac{\left(5x-2\right)\left(7x+6\right)}{35}$$

Cancel out 35, the greatest common factor in 35 and 35.

$$35x^{2}+16x-12=\left(5x-2\right)\left(7x+6\right)$$

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 35

$$x ^ 2 +\frac{16}{35}x -\frac{12}{35} = 0$$

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

$$r + s = -\frac{16}{35} rs = -\frac{12}{35}$$

Two numbers r and s sum up to -\frac{16}{35} exactly when the average of the two numbers is \frac{1}{2}*-\frac{16}{35} = -\frac{8}{35}. 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>

$$r = -\frac{8}{35} - u s = -\frac{8}{35} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{12}{35}

$$(-\frac{8}{35} - u) (-\frac{8}{35} + u) = -\frac{12}{35}$$

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

$$\frac{64}{1225} - u^2 = -\frac{12}{35}$$

Simplify the expression by subtracting \frac{64}{1225} on both sides

$$-u^2 = -\frac{12}{35}-\frac{64}{1225} = -\frac{484}{1225}$$

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

$$u^2 = \frac{484}{1225} u = \pm\sqrt{\frac{484}{1225}} = \pm \frac{22}{35}$$

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

$$r =-\frac{8}{35} - \frac{22}{35} = -0.857 s = -\frac{8}{35} + \frac{22}{35} = 0.400$$

Evaluate

$$\left(5x-2\right)\left(7x+6\right)$$