Solve r^2+7r-30

$$r ^ { 2 } + 7 r - 30$$

Factor

$$\left(r-3\right)\left(r+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br-30. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\left(-30\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite r^{2}+7r-30 as \left(r^{2}-3r\right)+\left(10r-30\right).

$$\left(r^{2}-3r\right)+\left(10r-30\right)$$

Factor out r in the first and 10 in the second group.

$$r\left(r-3\right)+10\left(r-3\right)$$

Factor out common term r-3 by using distributive property.

$$\left(r-3\right)\left(r+10\right)$$

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.

$$r^{2}+7r-30=0$$

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.

$$r=\frac{-7±\sqrt{7^{2}-4\left(-30\right)}}{2}$$

Square 7.

$$r=\frac{-7±\sqrt{49-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$r=\frac{-7±\sqrt{49+120}}{2}$$

Add 49 to 120.

$$r=\frac{-7±\sqrt{169}}{2}$$

Take the square root of 169.

$$r=\frac{-7±13}{2}$$

Now solve the equation r=\frac{-7±13}{2} when ± is plus. Add -7 to 13.

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

Divide 6 by 2.

$$r=3$$

Now solve the equation r=\frac{-7±13}{2} when ± is minus. Subtract 13 from -7.

$$r=-\frac{20}{2}$$

Divide -20 by 2.

$$r=-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -10 for x_{2}.

$$r^{2}+7r-30=\left(r-3\right)\left(r-\left(-10\right)\right)$$

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

$$r^{2}+7r-30=\left(r-3\right)\left(r+10\right)$$

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.

$$x ^ 2 +7x -30 = 0$$

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

$$r + s = -7 rs = -30$$

Two numbers r and s sum up to -7 exactly when the average of the two numbers is \frac{1}{2}*-7 = -\frac{7}{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>

$$r = -\frac{7}{2} - u s = -\frac{7}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -30

$$(-\frac{7}{2} - u) (-\frac{7}{2} + u) = -30$$

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

$$\frac{49}{4} - u^2 = -30$$

Simplify the expression by subtracting \frac{49}{4} on both sides

$$-u^2 = -30-\frac{49}{4} = -\frac{169}{4}$$

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.

$$r =-\frac{7}{2} - \frac{13}{2} = -10 s = -\frac{7}{2} + \frac{13}{2} = 3$$

Evaluate

$$\left(r-3\right)\left(r+10\right)$$