Solve 1-t-6t^2

$$1 - t - 6 t ^ { 2 }$$

Factor

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

Solution Steps

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

$$-6t^{2}-t+1$$

Factor the expression by grouping. First, the expression needs to be rewritten as -6t^{2}+at+bt+1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-6=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

Rewrite -6t^{2}-t+1 as \left(-6t^{2}+2t\right)+\left(-3t+1\right).

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

Factor out 2t in -6t^{2}+2t.

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

Factor out common term -3t+1 by using distributive property.

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

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.

$$-6t^{2}-t+1=0$$

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.

$$t=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$t=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-6\right)}$$

Add 1 to 24.

$$t=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-6\right)}$$

Take the square root of 25.

$$t=\frac{-\left(-1\right)±5}{2\left(-6\right)}$$

The opposite of -1 is 1.

$$t=\frac{1±5}{2\left(-6\right)}$$

Multiply 2 times -6.

$$t=\frac{1±5}{-12}$$

Now solve the equation t=\frac{1±5}{-12} when ± is plus. Add 1 to 5.

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

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{1±5}{-12} when ± is minus. Subtract 5 from 1.

$$t=-\frac{4}{-12}$$

Reduce the fraction \frac{-4}{-12} to lowest terms by extracting and canceling out 4.

$$t=\frac{1}{3}$$

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 \frac{1}{3} for x_{2}.

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

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

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

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

$$-6t^{2}-t+1=-6\times \frac{-2t-1}{-2}\left(t-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6t^{2}-t+1=-6\times \frac{-2t-1}{-2}\times \frac{-3t+1}{-3}$$

Multiply \frac{-2t-1}{-2} times \frac{-3t+1}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6t^{2}-t+1=-6\times \frac{\left(-2t-1\right)\left(-3t+1\right)}{-2\left(-3\right)}$$

Multiply -2 times -3.

$$-6t^{2}-t+1=-6\times \frac{\left(-2t-1\right)\left(-3t+1\right)}{6}$$

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

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

Evaluate

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