Solve r^2-2r-35

$$r ^ { 2 } - 2 r - 35$$

Factor

$$\left(r-7\right)\left(r+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br-35. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-35\right)=-35$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=-7 b=5$$

Rewrite r^{2}-2r-35 as \left(r^{2}-7r\right)+\left(5r-35\right).

$$\left(r^{2}-7r\right)+\left(5r-35\right)$$

Factor out r in the first and 5 in the second group.

$$r\left(r-7\right)+5\left(r-7\right)$$

Factor out common term r-7 by using distributive property.

$$\left(r-7\right)\left(r+5\right)$$

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.

$$r^{2}-2r-35=0$$

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.

$$r=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-35\right)}}{2}$$

Square -2.

$$r=\frac{-\left(-2\right)±\sqrt{4-4\left(-35\right)}}{2}$$

Multiply -4 times -35.

$$r=\frac{-\left(-2\right)±\sqrt{4+140}}{2}$$

Add 4 to 140.

$$r=\frac{-\left(-2\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$r=\frac{-\left(-2\right)±12}{2}$$

The opposite of -2 is 2.

$$r=\frac{2±12}{2}$$

Now solve the equation r=\frac{2±12}{2} when ± is plus. Add 2 to 12.

$$r=\frac{14}{2}$$

Divide 14 by 2.

$$r=7$$

Now solve the equation r=\frac{2±12}{2} when ± is minus. Subtract 12 from 2.

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

Divide -10 by 2.

$$r=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and -5 for x_{2}.

$$r^{2}-2r-35=\left(r-7\right)\left(r-\left(-5\right)\right)$$

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

$$r^{2}-2r-35=\left(r-7\right)\left(r+5\right)$$

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.

$$x ^ 2 -2x -35 = 0$$

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

$$r + s = 2 rs = -35$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -35

$$(1 - u) (1 + u) = -35$$

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

$$1 - u^2 = -35$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -35-1 = -36$$

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

$$u^2 = 36 u = \pm\sqrt{36} = \pm 6$$

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

$$r =1 - 6 = -5 s = 1 + 6 = 7$$

Evaluate

$$\left(r-7\right)\left(r+5\right)$$