Solve g(x)=x^2-3x-10

$$g ( x ) = x ^ { 2 } - 3 x - 10$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-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 negative, the negative number has greater absolute value than the positive. 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.

$$a=-5 b=2$$

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

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

Factor out x in the first and 2 in the second group.

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

Factor out common term x-5 by using distributive property.

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

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.

$$x^{2}-3x-10=0$$

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.

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

Square -3.

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

Multiply -4 times -10.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-3\right)±7}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±7}{2}$$

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

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

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

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

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

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

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

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

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.

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

Evaluate

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