Solve a^2-14a+45

$$a ^ { 2 } - 14 a + 45$$

Factor

$$\left(a-9\right)\left(a-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+45. To find p and q, set up a system to be solved.

$$p+q=-14 pq=1\times 45=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$p=-9 q=-5$$

Rewrite a^{2}-14a+45 as \left(a^{2}-9a\right)+\left(-5a+45\right).

$$\left(a^{2}-9a\right)+\left(-5a+45\right)$$

Factor out a in the first and -5 in the second group.

$$a\left(a-9\right)-5\left(a-9\right)$$

Factor out common term a-9 by using distributive property.

$$\left(a-9\right)\left(a-5\right)$$

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.

$$a^{2}-14a+45=0$$

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.

$$a=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 45}}{2}$$

Square -14.

$$a=\frac{-\left(-14\right)±\sqrt{196-4\times 45}}{2}$$

Multiply -4 times 45.

$$a=\frac{-\left(-14\right)±\sqrt{196-180}}{2}$$

Add 196 to -180.

$$a=\frac{-\left(-14\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$a=\frac{-\left(-14\right)±4}{2}$$

The opposite of -14 is 14.

$$a=\frac{14±4}{2}$$

Now solve the equation a=\frac{14±4}{2} when ± is plus. Add 14 to 4.

$$a=\frac{18}{2}$$

Divide 18 by 2.

$$a=9$$

Now solve the equation a=\frac{14±4}{2} when ± is minus. Subtract 4 from 14.

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

Divide 10 by 2.

$$a=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9 for x_{1} and 5 for x_{2}.

$$a^{2}-14a+45=\left(a-9\right)\left(a-5\right)$$

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.

$$x ^ 2 -14x +45 = 0$$

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

$$r + s = 14 rs = 45$$

Two numbers r and s sum up to 14 exactly when the average of the two numbers is \frac{1}{2}*14 = 7. 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>

$$r = 7 - u s = 7 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 45

$$(7 - u) (7 + u) = 45$$

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

$$49 - u^2 = 45$$

Simplify the expression by subtracting 49 on both sides

$$-u^2 = 45-49 = -4$$

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.

$$r =7 - 2 = 5 s = 7 + 2 = 9$$

Evaluate

$$\left(a-9\right)\left(a-5\right)$$