Factor the expression by grouping. First, the expression needs to be rewritten as $15t^{2}+at+bt-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=15\left(-2\right)=-30$$
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 $-30$.
$$1,-30$$ $$2,-15$$ $$3,-10$$ $$5,-6$$
Calculate the sum for each pair.
$$1-30=-29$$ $$2-15=-13$$ $$3-10=-7$$ $$5-6=-1$$
The solution is the pair that gives sum $-1$.
$$a=-6$$ $$b=5$$
Rewrite $15t^{2}-t-2$ as $\left(15t^{2}-6t\right)+\left(5t-2\right)$.
$$\left(15t^{2}-6t\right)+\left(5t-2\right)$$
Factor out $3t$ in $15t^{2}-6t$.
$$3t\left(5t-2\right)+5t-2$$
Factor out common term $5t-2$ by using distributive property.
$$\left(5t-2\right)\left(3t+1\right)$$
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 $ax^{2}+bx+c=0$.
$$15t^{2}-t-2=0$$
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.
Now solve the equation $t=\frac{1±11}{30}$ when $±$ is plus. Add $1$ to $11$.
$$t=\frac{12}{30}$$
Reduce the fraction $\frac{12}{30}$ to lowest terms by extracting and canceling out $6$.
$$t=\frac{2}{5}$$
Now solve the equation $t=\frac{1±11}{30}$ when $±$ is minus. Subtract $11$ from $1$.
$$t=-\frac{10}{30}$$
Reduce the fraction $\frac{-10}{30}$ to lowest terms by extracting and canceling out $10$.
$$t=-\frac{1}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{2}{5}$ for $x_{1}$ and $-\frac{1}{3}$ for $x_{2}$.
Multiply $\frac{5t-2}{5}$ times $\frac{3t+1}{3}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $15$
$$x ^ 2 -\frac{1}{15}x -\frac{2}{15} = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = \frac{1}{15} $$ $$ rs = -\frac{2}{15}$$
Two numbers $r$ and $s$ sum up to $\frac{1}{15}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{1}{15} = \frac{1}{30}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = \frac{1}{30} - u$$ $$s = \frac{1}{30} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -\frac{2}{15}$
