Solve 7x^2-2x-5

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

Factor

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

Solution Steps

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

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

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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.

$$a=-7 b=5$$

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

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

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

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

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

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

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.

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

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.

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

Square -2.

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

Multiply -4 times 7.

$$x=\frac{-\left(-2\right)±\sqrt{4-28\left(-5\right)}}{2\times 7}$$

Multiply -28 times -5.

$$x=\frac{-\left(-2\right)±\sqrt{4+140}}{2\times 7}$$

Add 4 to 140.

$$x=\frac{-\left(-2\right)±\sqrt{144}}{2\times 7}$$

Take the square root of 144.

$$x=\frac{-\left(-2\right)±12}{2\times 7}$$

The opposite of -2 is 2.

$$x=\frac{2±12}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{2±12}{14}$$

Now solve the equation x=\frac{2±12}{14} when ± is plus. Add 2 to 12.

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

Divide 14 by 14.

$$x=1$$

Now solve the equation x=\frac{2±12}{14} when ± is minus. Subtract 12 from 2.

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

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

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

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{5}{7} for x_{2}.

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

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

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

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

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

Cancel out 7, the greatest common factor in 7 and 7.

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

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 7

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

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

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

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

$$r = \frac{1}{7} - u s = \frac{1}{7} + u$$

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

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

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

$$\frac{1}{49} - u^2 = -\frac{5}{7}$$

Simplify the expression by subtracting \frac{1}{49} on both sides

$$-u^2 = -\frac{5}{7}-\frac{1}{49} = -\frac{36}{49}$$

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

$$u^2 = \frac{36}{49} u = \pm\sqrt{\frac{36}{49}} = \pm \frac{6}{7}$$

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

$$r =\frac{1}{7} - \frac{6}{7} = -0.714 s = \frac{1}{7} + \frac{6}{7} = 1$$

Evaluate

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