Solve f(t)=4+30t-16t^2

$$f ( t ) = 4 + 30 t - 16 t ^ { 2 }$$

Factor

$$2\left(2-t\right)\left(8t+1\right)$$

Solution Steps

Factor out 2.

$$2\left(2+15t-8t^{2}\right)$$

Consider 2+15t-8t^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-8t^{2}+15t+2$$

Factor the expression by grouping. First, the expression needs to be rewritten as -8t^{2}+at+bt+2. To find a and b, set up a system to be solved.

$$a+b=15 ab=-8\times 2=-16$$

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 -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 15.

$$a=16 b=-1$$

Rewrite -8t^{2}+15t+2 as \left(-8t^{2}+16t\right)+\left(-t+2\right).

$$\left(-8t^{2}+16t\right)+\left(-t+2\right)$$

Factor out 8t in -8t^{2}+16t.

$$8t\left(-t+2\right)-t+2$$

Factor out common term -t+2 by using distributive property.

$$\left(-t+2\right)\left(8t+1\right)$$

Rewrite the complete factored expression.

$$2\left(-t+2\right)\left(8t+1\right)$$

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.

$$-16t^{2}+30t+4=0$$

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.

$$t=\frac{-30±\sqrt{30^{2}-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Square 30.

$$t=\frac{-30±\sqrt{900-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-30±\sqrt{900+64\times 4}}{2\left(-16\right)}$$

Multiply 64 times 4.

$$t=\frac{-30±\sqrt{900+256}}{2\left(-16\right)}$$

Add 900 to 256.

$$t=\frac{-30±\sqrt{1156}}{2\left(-16\right)}$$

Take the square root of 1156.

$$t=\frac{-30±34}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-30±34}{-32}$$

Now solve the equation t=\frac{-30±34}{-32} when ± is plus. Add -30 to 34.

$$t=\frac{4}{-32}$$

Reduce the fraction \frac{4}{-32} to lowest terms by extracting and canceling out 4.

$$t=-\frac{1}{8}$$

Now solve the equation t=\frac{-30±34}{-32} when ± is minus. Subtract 34 from -30.

$$t=-\frac{64}{-32}$$

Divide -64 by -32.

$$t=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{8} for x_{1} and 2 for x_{2}.

$$-16t^{2}+30t+4=-16\left(t-\left(-\frac{1}{8}\right)\right)\left(t-2\right)$$

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

$$-16t^{2}+30t+4=-16\left(t+\frac{1}{8}\right)\left(t-2\right)$$

Add \frac{1}{8} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-16t^{2}+30t+4=-16\times \frac{-8t-1}{-8}\left(t-2\right)$$

Cancel out 8, the greatest common factor in -16 and 8.

$$-16t^{2}+30t+4=2\left(-8t-1\right)\left(t-2\right)$$

Evaluate

$$4+30t-16t^{2}$$