Solve 2x^2-3x+1)^2-22x^2+33x-1

$$2 x ^ { 2 } - 3 x + 1 ) ^ { 2 } - 22 x ^ { 2 } + 33 x - 1$$

Factor

$$\left(x-1\right)\left(2x-1\right)$$

Solution Steps

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

$$a+b=-3 ab=2\times 1=2$$

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.

$$a=-2 b=-1$$

Rewrite 2x^{2}-3x+1 as \left(2x^{2}-2x\right)+\left(-x+1\right).

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(x-1\right)-\left(x-1\right)$$

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

$$\left(x-1\right)\left(2x-1\right)$$

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.

$$2x^{2}-3x+1=0$$

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.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\times 2}$$

Add 9 to -8.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

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

The opposite of -3 is 3.

$$x=\frac{3±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±1}{4}$$

Now solve the equation x=\frac{3±1}{4} when ± is plus. Add 3 to 1.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{3±1}{4} when ± is minus. Subtract 1 from 3.

$$x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{2}$$

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}{2} for x_{2}.

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

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 2, the greatest common factor in 2 and 2.

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

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 2

$$x ^ 2 -\frac{3}{2}x +\frac{1}{2} = 0$$

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

$$r + s = \frac{3}{2} rs = \frac{1}{2}$$

Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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>

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = \frac{1}{2}$$

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

$$\frac{9}{16} - u^2 = \frac{1}{2}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = \frac{1}{2}-\frac{9}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =\frac{3}{4} - \frac{1}{4} = 0.500 s = \frac{3}{4} + \frac{1}{4} = 1$$

Evaluate

$$\left(x-1\right)\left(2x-1\right)$$