Solve f(x)=5x^2-6x+1

$$f ( x ) = 5 x ^ { 2 } - 6 x + 1$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-6 ab=5\times 1=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite 5x^{2}-6x+1 as \left(5x^{2}-5x\right)+\left(-x+1\right).

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

Factor out 5x in the first and -1 in the second group.

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

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

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

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.

$$5x^{2}-6x+1=0$$

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.

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

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20}}{2\times 5}$$

Add 36 to -20.

$$x=\frac{-\left(-6\right)±\sqrt{16}}{2\times 5}$$

Take the square root of 16.

$$x=\frac{-\left(-6\right)±4}{2\times 5}$$

The opposite of -6 is 6.

$$x=\frac{6±4}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{6±4}{10}$$

Now solve the equation x=\frac{6±4}{10} when ± is plus. Add 6 to 4.

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

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{6±4}{10} when ± is minus. Subtract 4 from 6.

$$x=\frac{2}{10}$$

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

$$x=\frac{1}{5}$$

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{1}{5} for x_{2}.

$$5x^{2}-6x+1=5\left(x-1\right)\left(x-\frac{1}{5}\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-6x+1=5\left(x-1\right)\times \frac{5x-1}{5}$$

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

$$5x^{2}-6x+1=\left(x-1\right)\left(5x-1\right)$$

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

$$x ^ 2 -\frac{6}{5}x +\frac{1}{5} = 0$$

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

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

Two numbers r and s sum up to \frac{6}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{6}{5} = \frac{3}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{5}

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

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

$$\frac{9}{25} - u^2 = \frac{1}{5}$$

Simplify the expression by subtracting \frac{9}{25} on both sides

$$-u^2 = \frac{1}{5}-\frac{9}{25} = -\frac{4}{25}$$

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

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

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

$$r =\frac{3}{5} - \frac{2}{5} = 0.200 s = \frac{3}{5} + \frac{2}{5} = 1$$

Evaluate

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