Solve r^2-17r+42

$$r ^ { 2 } - 17 r + 42$$

Factor

$$\left(r-14\right)\left(r-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br+42. To find a and b, set up a system to be solved.

$$a+b=-17 ab=1\times 42=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -17.

$$a=-14 b=-3$$

Rewrite r^{2}-17r+42 as \left(r^{2}-14r\right)+\left(-3r+42\right).

$$\left(r^{2}-14r\right)+\left(-3r+42\right)$$

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

$$r\left(r-14\right)-3\left(r-14\right)$$

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

$$\left(r-14\right)\left(r-3\right)$$

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.

$$r^{2}-17r+42=0$$

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.

$$r=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 42}}{2}$$

Square -17.

$$r=\frac{-\left(-17\right)±\sqrt{289-4\times 42}}{2}$$

Multiply -4 times 42.

$$r=\frac{-\left(-17\right)±\sqrt{289-168}}{2}$$

Add 289 to -168.

$$r=\frac{-\left(-17\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$r=\frac{-\left(-17\right)±11}{2}$$

The opposite of -17 is 17.

$$r=\frac{17±11}{2}$$

Now solve the equation r=\frac{17±11}{2} when ± is plus. Add 17 to 11.

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

Divide 28 by 2.

$$r=14$$

Now solve the equation r=\frac{17±11}{2} when ± is minus. Subtract 11 from 17.

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

Divide 6 by 2.

$$r=3$$

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 3 for x_{2}.

$$r^{2}-17r+42=\left(r-14\right)\left(r-3\right)$$

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.

$$x ^ 2 -17x +42 = 0$$

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

$$r + s = 17 rs = 42$$

Two numbers r and s sum up to 17 exactly when the average of the two numbers is \frac{1}{2}*17 = \frac{17}{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>

$$r = \frac{17}{2} - u s = \frac{17}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 42

$$(\frac{17}{2} - u) (\frac{17}{2} + u) = 42$$

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

$$\frac{289}{4} - u^2 = 42$$

Simplify the expression by subtracting \frac{289}{4} on both sides

$$-u^2 = 42-\frac{289}{4} = -\frac{121}{4}$$

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

$$u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{2}$$

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

$$r =\frac{17}{2} - \frac{11}{2} = 3 s = \frac{17}{2} + \frac{11}{2} = 14$$

Evaluate

$$\left(r-14\right)\left(r-3\right)$$