Solve 28x^2+2x-6

$$28 x ^ { 2 } + 2 x - 6$$

Factor

$$2\left(7x-3\right)\left(2x+1\right)$$

Solution Steps

Factor out 2.

$$2\left(14x^{2}+x-3\right)$$

Consider 14x^{2}+x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 14x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=1 ab=14\left(-3\right)=-42$$

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 -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=-6 b=7$$

Rewrite 14x^{2}+x-3 as \left(14x^{2}-6x\right)+\left(7x-3\right).

$$\left(14x^{2}-6x\right)+\left(7x-3\right)$$

Factor out 2x in 14x^{2}-6x.

$$2x\left(7x-3\right)+7x-3$$

Factor out common term 7x-3 by using distributive property.

$$\left(7x-3\right)\left(2x+1\right)$$

Rewrite the complete factored expression.

$$2\left(7x-3\right)\left(2x+1\right)$$

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.

$$28x^{2}+2x-6=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 28\left(-6\right)}}{2\times 28}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 28\left(-6\right)}}{2\times 28}$$

Multiply -4 times 28.

$$x=\frac{-2±\sqrt{4-112\left(-6\right)}}{2\times 28}$$

Multiply -112 times -6.

$$x=\frac{-2±\sqrt{4+672}}{2\times 28}$$

Add 4 to 672.

$$x=\frac{-2±\sqrt{676}}{2\times 28}$$

Take the square root of 676.

$$x=\frac{-2±26}{2\times 28}$$

Multiply 2 times 28.

$$x=\frac{-2±26}{56}$$

Now solve the equation x=\frac{-2±26}{56} when ± is plus. Add -2 to 26.

$$x=\frac{24}{56}$$

Reduce the fraction \frac{24}{56} to lowest terms by extracting and canceling out 8.

$$x=\frac{3}{7}$$

Now solve the equation x=\frac{-2±26}{56} when ± is minus. Subtract 26 from -2.

$$x=-\frac{28}{56}$$

Reduce the fraction \frac{-28}{56} to lowest terms by extracting and canceling out 28.

$$x=-\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{7} for x_{1} and -\frac{1}{2} for x_{2}.

$$28x^{2}+2x-6=28\left(x-\frac{3}{7}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$28x^{2}+2x-6=28\left(x-\frac{3}{7}\right)\left(x+\frac{1}{2}\right)$$

Subtract \frac{3}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$28x^{2}+2x-6=28\times \frac{7x-3}{7}\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$28x^{2}+2x-6=28\times \frac{7x-3}{7}\times \frac{2x+1}{2}$$

Multiply \frac{7x-3}{7} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$28x^{2}+2x-6=28\times \frac{\left(7x-3\right)\left(2x+1\right)}{7\times 2}$$

Multiply 7 times 2.

$$28x^{2}+2x-6=28\times \frac{\left(7x-3\right)\left(2x+1\right)}{14}$$

Cancel out 14, the greatest common factor in 28 and 14.

$$28x^{2}+2x-6=2\left(7x-3\right)\left(2x+1\right)$$

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 28

$$x ^ 2 +\frac{1}{14}x -\frac{3}{14} = 0$$

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

$$r + s = -\frac{1}{14} rs = -\frac{3}{14}$$

Two numbers r and s sum up to -\frac{1}{14} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{14} = -\frac{1}{28}. 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>

$$r = -\frac{1}{28} - u s = -\frac{1}{28} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{14}

$$(-\frac{1}{28} - u) (-\frac{1}{28} + u) = -\frac{3}{14}$$

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

$$\frac{1}{784} - u^2 = -\frac{3}{14}$$

Simplify the expression by subtracting \frac{1}{784} on both sides

$$-u^2 = -\frac{3}{14}-\frac{1}{784} = -\frac{169}{784}$$

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

$$u^2 = \frac{169}{784} u = \pm\sqrt{\frac{169}{784}} = \pm \frac{13}{28}$$

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

$$r =-\frac{1}{28} - \frac{13}{28} = -0.500 s = -\frac{1}{28} + \frac{13}{28} = 0.429$$

Evaluate

$$2\left(14x^{2}+x-3\right)$$