Solve 8j^2+20j-28

$$8 j ^ { 2 } + 20 j - 28$$

Factor

$$4\left(j-1\right)\left(2j+7\right)$$

Solution Steps

Factor out 4.

$$4\left(2j^{2}+5j-7\right)$$

Consider 2j^{2}+5j-7. Factor the expression by grouping. First, the expression needs to be rewritten as 2j^{2}+aj+bj-7. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\left(-7\right)=-14$$

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 -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

Rewrite 2j^{2}+5j-7 as \left(2j^{2}-2j\right)+\left(7j-7\right).

$$\left(2j^{2}-2j\right)+\left(7j-7\right)$$

Factor out 2j in the first and 7 in the second group.

$$2j\left(j-1\right)+7\left(j-1\right)$$

Factor out common term j-1 by using distributive property.

$$\left(j-1\right)\left(2j+7\right)$$

Rewrite the complete factored expression.

$$4\left(j-1\right)\left(2j+7\right)$$

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.

$$8j^{2}+20j-28=0$$

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.

$$j=\frac{-20±\sqrt{20^{2}-4\times 8\left(-28\right)}}{2\times 8}$$

Square 20.

$$j=\frac{-20±\sqrt{400-4\times 8\left(-28\right)}}{2\times 8}$$

Multiply -4 times 8.

$$j=\frac{-20±\sqrt{400-32\left(-28\right)}}{2\times 8}$$

Multiply -32 times -28.

$$j=\frac{-20±\sqrt{400+896}}{2\times 8}$$

Add 400 to 896.

$$j=\frac{-20±\sqrt{1296}}{2\times 8}$$

Take the square root of 1296.

$$j=\frac{-20±36}{2\times 8}$$

Multiply 2 times 8.

$$j=\frac{-20±36}{16}$$

Now solve the equation j=\frac{-20±36}{16} when ± is plus. Add -20 to 36.

$$j=\frac{16}{16}$$

Divide 16 by 16.

$$j=1$$

Now solve the equation j=\frac{-20±36}{16} when ± is minus. Subtract 36 from -20.

$$j=-\frac{56}{16}$$

Reduce the fraction \frac{-56}{16} to lowest terms by extracting and canceling out 8.

$$j=-\frac{7}{2}$$

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

$$8j^{2}+20j-28=8\left(j-1\right)\left(j-\left(-\frac{7}{2}\right)\right)$$

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

$$8j^{2}+20j-28=8\left(j-1\right)\left(j+\frac{7}{2}\right)$$

Add \frac{7}{2} to j by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8j^{2}+20j-28=8\left(j-1\right)\times \frac{2j+7}{2}$$

Cancel out 2, the greatest common factor in 8 and 2.

$$8j^{2}+20j-28=4\left(j-1\right)\left(2j+7\right)$$

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 8

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

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

$$r + s = -\frac{5}{2} rs = -\frac{7}{2}$$

Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{4}. 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>

$$r = -\frac{5}{4} - u s = -\frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{2}

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = -\frac{7}{2}$$

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

$$\frac{25}{16} - u^2 = -\frac{7}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -\frac{7}{2}-\frac{25}{16} = -\frac{81}{16}$$

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

$$u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}$$

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

$$r =-\frac{5}{4} - \frac{9}{4} = -3.500 s = -\frac{5}{4} + \frac{9}{4} = 1$$

Evaluate

$$4\left(j-1\right)\left(2j+7\right)$$