Solve 33t-15-6t^2

$$33 t - 15 - 6 t ^ { 2 }$$

Factor

$$3\left(5-t\right)\left(2t-1\right)$$

Solution Steps

Factor out 3.

$$3\left(11t-5-2t^{2}\right)$$

Consider 11t-5-2t^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2t^{2}+11t-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as -2t^{2}+at+bt-5. To find a and b, set up a system to be solved.

$$a+b=11 ab=-2\left(-5\right)=10$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 11.

$$a=10 b=1$$

Rewrite -2t^{2}+11t-5 as \left(-2t^{2}+10t\right)+\left(t-5\right).

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

Factor out 2t in the first and -1 in the second group.

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

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

$$\left(-t+5\right)\left(2t-1\right)$$

Rewrite the complete factored expression.

$$3\left(-t+5\right)\left(2t-1\right)$$

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.

$$-6t^{2}+33t-15=0$$

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.

$$t=\frac{-33±\sqrt{33^{2}-4\left(-6\right)\left(-15\right)}}{2\left(-6\right)}$$

Square 33.

$$t=\frac{-33±\sqrt{1089-4\left(-6\right)\left(-15\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$t=\frac{-33±\sqrt{1089+24\left(-15\right)}}{2\left(-6\right)}$$

Multiply 24 times -15.

$$t=\frac{-33±\sqrt{1089-360}}{2\left(-6\right)}$$

Add 1089 to -360.

$$t=\frac{-33±\sqrt{729}}{2\left(-6\right)}$$

Take the square root of 729.

$$t=\frac{-33±27}{2\left(-6\right)}$$

Multiply 2 times -6.

$$t=\frac{-33±27}{-12}$$

Now solve the equation t=\frac{-33±27}{-12} when ± is plus. Add -33 to 27.

$$t=-\frac{6}{-12}$$

Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.

$$t=\frac{1}{2}$$

Now solve the equation t=\frac{-33±27}{-12} when ± is minus. Subtract 27 from -33.

$$t=-\frac{60}{-12}$$

Divide -60 by -12.

$$t=5$$

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

$$-6t^{2}+33t-15=-6\left(t-\frac{1}{2}\right)\left(t-5\right)$$

Subtract \frac{1}{2} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6t^{2}+33t-15=-6\times \frac{-2t+1}{-2}\left(t-5\right)$$

Cancel out 2, the greatest common factor in -6 and 2.

$$-6t^{2}+33t-15=3\left(-2t+1\right)\left(t-5\right)$$

Evaluate

$$-6t^{2}+33t-15$$