Solve x^2-19x+48=

$$x ^ { 2 } - 19 x + 48 =$$

Factor

$$\left(x-16\right)\left(x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+48. To find a and b, set up a system to be solved.

$$a+b=-19 ab=1\times 48=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -19.

$$a=-16 b=-3$$

Rewrite x^{2}-19x+48 as \left(x^{2}-16x\right)+\left(-3x+48\right).

$$\left(x^{2}-16x\right)+\left(-3x+48\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-16\right)-3\left(x-16\right)$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(x-3\right)$$

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.

$$x^{2}-19x+48=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 48}}{2}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 48}}{2}$$

Multiply -4 times 48.

$$x=\frac{-\left(-19\right)±\sqrt{361-192}}{2}$$

Add 361 to -192.

$$x=\frac{-\left(-19\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-\left(-19\right)±13}{2}$$

The opposite of -19 is 19.

$$x=\frac{19±13}{2}$$

Now solve the equation x=\frac{19±13}{2} when ± is plus. Add 19 to 13.

$$x=\frac{32}{2}$$

Divide 32 by 2.

$$x=16$$

Now solve the equation x=\frac{19±13}{2} when ± is minus. Subtract 13 from 19.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 16 for x_{1} and 3 for x_{2}.

$$x^{2}-19x+48=\left(x-16\right)\left(x-3\right)$$

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.

$$x ^ 2 -19x +48 = 0$$

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

$$r + s = 19 rs = 48$$

Two numbers r and s sum up to 19 exactly when the average of the two numbers is \frac{1}{2}*19 = \frac{19}{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>

$$r = \frac{19}{2} - u s = \frac{19}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 48

$$(\frac{19}{2} - u) (\frac{19}{2} + u) = 48$$

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

$$\frac{361}{4} - u^2 = 48$$

Simplify the expression by subtracting \frac{361}{4} on both sides

$$-u^2 = 48-\frac{361}{4} = -\frac{169}{4}$$

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

$$u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}$$

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

$$r =\frac{19}{2} - \frac{13}{2} = 3 s = \frac{19}{2} + \frac{13}{2} = 16$$

Evaluate

$$\left(x-16\right)\left(x-3\right)$$