Solve a^2-6a+5

$$a ^ { 2 } - 6 a + 5$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+5. To find p and q, set up a system to be solved.

$$p+q=-6 pq=1\times 5=5$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. The only such pair is the system solution.

$$p=-5 q=-1$$

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

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

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

$$a\left(a-5\right)-\left(a-5\right)$$

Factor out common term a-5 by using distributive property.

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

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.

$$a^{2}-6a+5=0$$

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.

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

Square -6.

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

Multiply -4 times 5.

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

Add 36 to -20.

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

Take the square root of 16.

$$a=\frac{-\left(-6\right)±4}{2}$$

The opposite of -6 is 6.

$$a=\frac{6±4}{2}$$

Now solve the equation a=\frac{6±4}{2} when ± is plus. Add 6 to 4.

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

Divide 10 by 2.

$$a=5$$

Now solve the equation a=\frac{6±4}{2} when ± is minus. Subtract 4 from 6.

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

Divide 2 by 2.

$$a=1$$

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

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

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.

$$x ^ 2 -6x +5 = 0$$

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

$$r + s = 6 rs = 5$$

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

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(3 - u) (3 + u) = 5$$

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

$$9 - u^2 = 5$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 5-9 = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =3 - 2 = 1 s = 3 + 2 = 5$$

Evaluate

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