Solve a^2-4a-45=

$$a ^ { 2 } - 4 a - 45 =$$

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.

$$p+q=-4 pq=1\left(-45\right)=-45$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -45.

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

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -4.

$$p=-9 q=5$$

Rewrite a^{2}-4a-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.

$$a^{2}-4a-45=0$$

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.

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

Square -4.

$$a=\frac{-\left(-4\right)±\sqrt{16-4\left(-45\right)}}{2}$$

Multiply -4 times -45.

$$a=\frac{-\left(-4\right)±\sqrt{16+180}}{2}$$

Add 16 to 180.

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

Take the square root of 196.

$$a=\frac{-\left(-4\right)±14}{2}$$

The opposite of -4 is 4.

$$a=\frac{4±14}{2}$$

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

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

Divide 18 by 2.

$$a=9$$

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

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

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}.

$$a^{2}-4a-45=\left(a-9\right)\left(a-\left(-5\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$a^{2}-4a-45=\left(a-9\right)\left(a+5\right)$$

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.

$$x ^ 2 -4x -45 = 0$$

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

$$r + s = 4 rs = -45$$

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

$$r = 2 - u s = 2 + u$$

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

$$(2 - u) (2 + u) = -45$$

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

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

Simplify the expression by subtracting 4 on both sides

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

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

$$u^2 = 49 u = \pm\sqrt{49} = \pm 7$$

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

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

Evaluate

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