Solve r^2+3r-10

$$r ^ { 2 } + 3 r - 10$$

Factor

$$\left(r-2\right)\left(r+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-10\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite r^{2}+3r-10 as \left(r^{2}-2r\right)+\left(5r-10\right).

$$\left(r^{2}-2r\right)+\left(5r-10\right)$$

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

$$r\left(r-2\right)+5\left(r-2\right)$$

Factor out common term r-2 by using distributive property.

$$\left(r-2\right)\left(r+5\right)$$

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.

$$r^{2}+3r-10=0$$

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.

$$r=\frac{-3±\sqrt{3^{2}-4\left(-10\right)}}{2}$$

Square 3.

$$r=\frac{-3±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$r=\frac{-3±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$r=\frac{-3±\sqrt{49}}{2}$$

Take the square root of 49.

$$r=\frac{-3±7}{2}$$

Now solve the equation r=\frac{-3±7}{2} when ± is plus. Add -3 to 7.

$$r=\frac{4}{2}$$

Divide 4 by 2.

$$r=2$$

Now solve the equation r=\frac{-3±7}{2} when ± is minus. Subtract 7 from -3.

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

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 2 for x_{1} and -5 for x_{2}.

$$r^{2}+3r-10=\left(r-2\right)\left(r-\left(-5\right)\right)$$

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

$$r^{2}+3r-10=\left(r-2\right)\left(r+5\right)$$

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.

$$x ^ 2 +3x -10 = 0$$

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

$$r + s = -3 rs = -10$$

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{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>

$$r = -\frac{3}{2} - u s = -\frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -10$$

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

$$\frac{9}{4} - u^2 = -10$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -10-\frac{9}{4} = -\frac{49}{4}$$

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

$$u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}$$

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

$$r =-\frac{3}{2} - \frac{7}{2} = -5 s = -\frac{3}{2} + \frac{7}{2} = 2$$

Evaluate

$$\left(r-2\right)\left(r+5\right)$$