Solve a^2-18a+56

$${ a }^{ 2 } -18a+56$$

Factor

$$\left(a-14\right)\left(a-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+56. To find p and q, set up a system to be solved.

$$p+q=-18 pq=1\times 56=56$$

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 56.

$$-1,-56 -2,-28 -4,-14 -7,-8$$

Calculate the sum for each pair.

$$-1-56=-57 -2-28=-30 -4-14=-18 -7-8=-15$$

The solution is the pair that gives sum -18.

$$p=-14 q=-4$$

Rewrite a^{2}-18a+56 as \left(a^{2}-14a\right)+\left(-4a+56\right).

$$\left(a^{2}-14a\right)+\left(-4a+56\right)$$

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

$$a\left(a-14\right)-4\left(a-14\right)$$

Factor out common term a-14 by using distributive property.

$$\left(a-14\right)\left(a-4\right)$$

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.

$$a^{2}-18a+56=0$$

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.

$$a=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 56}}{2}$$

Square -18.

$$a=\frac{-\left(-18\right)±\sqrt{324-4\times 56}}{2}$$

Multiply -4 times 56.

$$a=\frac{-\left(-18\right)±\sqrt{324-224}}{2}$$

Add 324 to -224.

$$a=\frac{-\left(-18\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$a=\frac{-\left(-18\right)±10}{2}$$

The opposite of -18 is 18.

$$a=\frac{18±10}{2}$$

Now solve the equation a=\frac{18±10}{2} when ± is plus. Add 18 to 10.

$$a=\frac{28}{2}$$

Divide 28 by 2.

$$a=14$$

Now solve the equation a=\frac{18±10}{2} when ± is minus. Subtract 10 from 18.

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

Divide 8 by 2.

$$a=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 14 for x_{1} and 4 for x_{2}.

$$a^{2}-18a+56=\left(a-14\right)\left(a-4\right)$$

Evaluate

$$\left(a-14\right)\left(a-4\right)$$