Solve a^2-10a+24

$$a ^ { 2 } - 10 a + 24$$

Factor

$$\left(a-6\right)\left(a-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+24. To find p and q, set up a system to be solved.

$$p+q=-10 pq=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$p=-6 q=-4$$

Rewrite a^{2}-10a+24 as \left(a^{2}-6a\right)+\left(-4a+24\right).

$$\left(a^{2}-6a\right)+\left(-4a+24\right)$$

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

$$a\left(a-6\right)-4\left(a-6\right)$$

Factor out common term a-6 by using distributive property.

$$\left(a-6\right)\left(a-4\right)$$

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.

$$a^{2}-10a+24=0$$

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.

$$a=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 24}}{2}$$

Square -10.

$$a=\frac{-\left(-10\right)±\sqrt{100-4\times 24}}{2}$$

Multiply -4 times 24.

$$a=\frac{-\left(-10\right)±\sqrt{100-96}}{2}$$

Add 100 to -96.

$$a=\frac{-\left(-10\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$a=\frac{-\left(-10\right)±2}{2}$$

The opposite of -10 is 10.

$$a=\frac{10±2}{2}$$

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

$$a=\frac{12}{2}$$

Divide 12 by 2.

$$a=6$$

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

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

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 6 for x_{1} and 4 for x_{2}.

$$a^{2}-10a+24=\left(a-6\right)\left(a-4\right)$$

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.

$$x ^ 2 -10x +24 = 0$$

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

$$r + s = 10 rs = 24$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>

$$r = 5 - u s = 5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 24

$$(5 - u) (5 + u) = 24$$

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

$$25 - u^2 = 24$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 24-25 = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =5 - 1 = 4 s = 5 + 1 = 6$$

Evaluate

$$\left(a-6\right)\left(a-4\right)$$