Solve 3t-28+t^2

$$3 t - 28 + t ^ { 2 }$$

Factor

$$\left(t-4\right)\left(t+7\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$t^{2}+3t-28$$

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-28. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-28\right)=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite t^{2}+3t-28 as \left(t^{2}-4t\right)+\left(7t-28\right).

$$\left(t^{2}-4t\right)+\left(7t-28\right)$$

Factor out t in the first and 7 in the second group.

$$t\left(t-4\right)+7\left(t-4\right)$$

Factor out common term t-4 by using distributive property.

$$\left(t-4\right)\left(t+7\right)$$

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.

$$t^{2}+3t-28=0$$

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.

$$t=\frac{-3±\sqrt{3^{2}-4\left(-28\right)}}{2}$$

Square 3.

$$t=\frac{-3±\sqrt{9-4\left(-28\right)}}{2}$$

Multiply -4 times -28.

$$t=\frac{-3±\sqrt{9+112}}{2}$$

Add 9 to 112.

$$t=\frac{-3±\sqrt{121}}{2}$$

Take the square root of 121.

$$t=\frac{-3±11}{2}$$

Now solve the equation t=\frac{-3±11}{2} when ± is plus. Add -3 to 11.

$$t=\frac{8}{2}$$

Divide 8 by 2.

$$t=4$$

Now solve the equation t=\frac{-3±11}{2} when ± is minus. Subtract 11 from -3.

$$t=-\frac{14}{2}$$

Divide -14 by 2.

$$t=-7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -7 for x_{2}.

$$t^{2}+3t-28=\left(t-4\right)\left(t-\left(-7\right)\right)$$

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

$$t^{2}+3t-28=\left(t-4\right)\left(t+7\right)$$

Evaluate

$$\left(t-4\right)\left(t+7\right)$$