Solve 4a^2-5a+1

$$4 a ^ { 2 } - 5 a + 1$$

Factor

$$\left(a-1\right)\left(4a-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4a^{2}+pa+qa+1. To find p and q, set up a system to be solved.

$$p+q=-5 pq=4\times 1=4$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$p=-4 q=-1$$

Rewrite 4a^{2}-5a+1 as \left(4a^{2}-4a\right)+\left(-a+1\right).

$$\left(4a^{2}-4a\right)+\left(-a+1\right)$$

Factor out 4a in the first and -1 in the second group.

$$4a\left(a-1\right)-\left(a-1\right)$$

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

$$\left(a-1\right)\left(4a-1\right)$$

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.

$$4a^{2}-5a+1=0$$

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.

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

Square -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$a=\frac{-\left(-5\right)±\sqrt{25-16}}{2\times 4}$$

Add 25 to -16.

$$a=\frac{-\left(-5\right)±\sqrt{9}}{2\times 4}$$

Take the square root of 9.

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

The opposite of -5 is 5.

$$a=\frac{5±3}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{5±3}{8}$$

Now solve the equation a=\frac{5±3}{8} when ± is plus. Add 5 to 3.

$$a=\frac{8}{8}$$

Divide 8 by 8.

$$a=1$$

Now solve the equation a=\frac{5±3}{8} when ± is minus. Subtract 3 from 5.

$$a=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$a=\frac{1}{4}$$

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}{4} for x_{2}.

$$4a^{2}-5a+1=4\left(a-1\right)\left(a-\frac{1}{4}\right)$$

Subtract \frac{1}{4} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4a^{2}-5a+1=4\left(a-1\right)\times \frac{4a-1}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4a^{2}-5a+1=\left(a-1\right)\left(4a-1\right)$$

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 4

$$x ^ 2 -\frac{5}{4}x +\frac{1}{4} = 0$$

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

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

Two numbers r and s sum up to \frac{5}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{4} = \frac{5}{8}. 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>

$$r = \frac{5}{8} - u s = \frac{5}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{4}

$$(\frac{5}{8} - u) (\frac{5}{8} + u) = \frac{1}{4}$$

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

$$\frac{25}{64} - u^2 = \frac{1}{4}$$

Simplify the expression by subtracting \frac{25}{64} on both sides

$$-u^2 = \frac{1}{4}-\frac{25}{64} = -\frac{9}{64}$$

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

$$u^2 = \frac{9}{64} u = \pm\sqrt{\frac{9}{64}} = \pm \frac{3}{8}$$

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

$$r =\frac{5}{8} - \frac{3}{8} = 0.250 s = \frac{5}{8} + \frac{3}{8} = 1$$

Evaluate

$$\left(a-1\right)\left(4a-1\right)$$