Solve 5x^2+2x-7

$$5 x ^ { 2 } + 2 x - 7$$

Factor

$$\left(x-1\right)\left(5x+7\right)$$

Solution Steps

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

$$a+b=2 ab=5\left(-7\right)=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=-5 b=7$$

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

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

Factor out 5x in the first and 7 in the second group.

$$5x\left(x-1\right)+7\left(x-1\right)$$

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

$$\left(x-1\right)\left(5x+7\right)$$

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.

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

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 5\left(-7\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-2±\sqrt{4-20\left(-7\right)}}{2\times 5}$$

Multiply -20 times -7.

$$x=\frac{-2±\sqrt{4+140}}{2\times 5}$$

Add 4 to 140.

$$x=\frac{-2±\sqrt{144}}{2\times 5}$$

Take the square root of 144.

$$x=\frac{-2±12}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-2±12}{10}$$

Now solve the equation x=\frac{-2±12}{10} when ± is plus. Add -2 to 12.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{-2±12}{10} when ± is minus. Subtract 12 from -2.

$$x=-\frac{14}{10}$$

Reduce the fraction \frac{-14}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{5}$$

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}{5} for x_{2}.

$$5x^{2}+2x-7=5\left(x-1\right)\left(x-\left(-\frac{7}{5}\right)\right)$$

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

$$5x^{2}+2x-7=5\left(x-1\right)\left(x+\frac{7}{5}\right)$$

Add \frac{7}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+2x-7=5\left(x-1\right)\times \frac{5x+7}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5x^{2}+2x-7=\left(x-1\right)\left(5x+7\right)$$

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 5

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

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

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

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

$$r = -\frac{1}{5} - u s = -\frac{1}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{5}

$$(-\frac{1}{5} - u) (-\frac{1}{5} + u) = -\frac{7}{5}$$

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

$$\frac{1}{25} - u^2 = -\frac{7}{5}$$

Simplify the expression by subtracting \frac{1}{25} on both sides

$$-u^2 = -\frac{7}{5}-\frac{1}{25} = -\frac{36}{25}$$

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

$$u^2 = \frac{36}{25} u = \pm\sqrt{\frac{36}{25}} = \pm \frac{6}{5}$$

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

$$r =-\frac{1}{5} - \frac{6}{5} = -1.400 s = -\frac{1}{5} + \frac{6}{5} = 1$$

Evaluate

$$\left(x-1\right)\left(5x+7\right)$$