Solve f(x)=5x^2-2x-3

$$f ( x ) = 5 x ^ { 2 } - 2 x - 3$$

Factor

$$\left(x-1\right)\left(5x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=5\left(-3\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite 5x^{2}-2x-3 as \left(5x^{2}-5x\right)+\left(3x-3\right).

$$\left(5x^{2}-5x\right)+\left(3x-3\right)$$

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

$$5x\left(x-1\right)+3\left(x-1\right)$$

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

$$\left(x-1\right)\left(5x+3\right)$$

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.

$$5x^{2}-2x-3=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-2\right)±\sqrt{4-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+60}}{2\times 5}$$

Add 4 to 60.

$$x=\frac{-\left(-2\right)±\sqrt{64}}{2\times 5}$$

Take the square root of 64.

$$x=\frac{-\left(-2\right)±8}{2\times 5}$$

The opposite of -2 is 2.

$$x=\frac{2±8}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{2±8}{10}$$

Now solve the equation x=\frac{2±8}{10} when ± is plus. Add 2 to 8.

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

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{2±8}{10} when ± is minus. Subtract 8 from 2.

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

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

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

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{3}{5} for x_{2}.

$$5x^{2}-2x-3=5\left(x-1\right)\left(x-\left(-\frac{3}{5}\right)\right)$$

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

$$5x^{2}-2x-3=5\left(x-1\right)\left(x+\frac{3}{5}\right)$$

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

$$5x^{2}-2x-3=5\left(x-1\right)\times \frac{5x+3}{5}$$

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

$$5x^{2}-2x-3=\left(x-1\right)\left(5x+3\right)$$

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

$$x ^ 2 -\frac{2}{5}x -\frac{3}{5} = 0$$

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

$$r + s = \frac{2}{5} rs = -\frac{3}{5}$$

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{3}{5}

$$(\frac{1}{5} - u) (\frac{1}{5} + u) = -\frac{3}{5}$$

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

$$\frac{1}{25} - u^2 = -\frac{3}{5}$$

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

$$-u^2 = -\frac{3}{5}-\frac{1}{25} = -\frac{16}{25}$$

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

$$u^2 = \frac{16}{25} u = \pm\sqrt{\frac{16}{25}} = \pm \frac{4}{5}$$

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

$$r =\frac{1}{5} - \frac{4}{5} = -0.600 s = \frac{1}{5} + \frac{4}{5} = 1$$

Evaluate

$$\left(x-1\right)\left(5x+3\right)$$