Solve (t^2-3t-10)

$$( t ^ { 2 } - 3 t - 10 )$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-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 t^{2}-3t-10 as \left(t^{2}-5t\right)+\left(2t-10\right).

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

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

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

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

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

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.

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

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.

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

Square -3.

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

Multiply -4 times -10.

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

Add 9 to 40.

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

Take the square root of 49.

$$t=\frac{-\left(-3\right)±7}{2}$$

The opposite of -3 is 3.

$$t=\frac{3±7}{2}$$

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

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

Divide 10 by 2.

$$t=5$$

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

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

Divide -4 by 2.

$$t=-2$$

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}.

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

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

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

Evaluate

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