Solve 5s^2-7s-6

$$5 s ^ { 2 } - 7 s - 6$$

Factor

$$\left(s-2\right)\left(5s+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5s^{2}+as+bs-6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=5\left(-6\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=-10 b=3$$

Rewrite 5s^{2}-7s-6 as \left(5s^{2}-10s\right)+\left(3s-6\right).

$$\left(5s^{2}-10s\right)+\left(3s-6\right)$$

Factor out 5s in the first and 3 in the second group.

$$5s\left(s-2\right)+3\left(s-2\right)$$

Factor out common term s-2 by using distributive property.

$$\left(s-2\right)\left(5s+3\right)$$

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.

$$5s^{2}-7s-6=0$$

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.

$$s=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5\left(-6\right)}}{2\times 5}$$

Square -7.

$$s=\frac{-\left(-7\right)±\sqrt{49-4\times 5\left(-6\right)}}{2\times 5}$$

Multiply -4 times 5.

$$s=\frac{-\left(-7\right)±\sqrt{49-20\left(-6\right)}}{2\times 5}$$

Multiply -20 times -6.

$$s=\frac{-\left(-7\right)±\sqrt{49+120}}{2\times 5}$$

Add 49 to 120.

$$s=\frac{-\left(-7\right)±\sqrt{169}}{2\times 5}$$

Take the square root of 169.

$$s=\frac{-\left(-7\right)±13}{2\times 5}$$

The opposite of -7 is 7.

$$s=\frac{7±13}{2\times 5}$$

Multiply 2 times 5.

$$s=\frac{7±13}{10}$$

Now solve the equation s=\frac{7±13}{10} when ± is plus. Add 7 to 13.

$$s=\frac{20}{10}$$

Divide 20 by 10.

$$s=2$$

Now solve the equation s=\frac{7±13}{10} when ± is minus. Subtract 13 from 7.

$$s=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

$$s=-\frac{3}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -\frac{3}{5} for x_{2}.

$$5s^{2}-7s-6=5\left(s-2\right)\left(s-\left(-\frac{3}{5}\right)\right)$$

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

$$5s^{2}-7s-6=5\left(s-2\right)\left(s+\frac{3}{5}\right)$$

Add \frac{3}{5} to s by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5s^{2}-7s-6=5\left(s-2\right)\times \frac{5s+3}{5}$$

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

$$5s^{2}-7s-6=\left(s-2\right)\left(5s+3\right)$$

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

$$x ^ 2 -\frac{7}{5}x -\frac{6}{5} = 0$$

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

$$r + s = \frac{7}{5} rs = -\frac{6}{5}$$

Two numbers r and s sum up to \frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{5} = \frac{7}{10}. 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>

$$r = \frac{7}{10} - u s = \frac{7}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{6}{5}

$$(\frac{7}{10} - u) (\frac{7}{10} + u) = -\frac{6}{5}$$

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

$$\frac{49}{100} - u^2 = -\frac{6}{5}$$

Simplify the expression by subtracting \frac{49}{100} on both sides

$$-u^2 = -\frac{6}{5}-\frac{49}{100} = -\frac{169}{100}$$

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

$$u^2 = \frac{169}{100} u = \pm\sqrt{\frac{169}{100}} = \pm \frac{13}{10}$$

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

$$r =\frac{7}{10} - \frac{13}{10} = -0.600 s = \frac{7}{10} + \frac{13}{10} = 2$$

Evaluate

$$\left(s-2\right)\left(5s+3\right)$$