Consider $10t^{2}+13t-3$. Factor the expression by grouping. First, the expression needs to be rewritten as $10t^{2}+at+bt-3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=13$$ $$ab=10\left(-3\right)=-30$$
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 $-30$.
$$-1,30$$ $$-2,15$$ $$-3,10$$ $$-5,6$$
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 $13$.
$$a=-2$$ $$b=15$$
Rewrite $10t^{2}+13t-3$ as $\left(10t^{2}-2t\right)+\left(15t-3\right)$.
$$\left(10t^{2}-2t\right)+\left(15t-3\right)$$
Factor out $2t$ in the first and $3$ in the second group.
$$2t\left(5t-1\right)+3\left(5t-1\right)$$
Factor out common term $5t-1$ by using distributive property.
$$\left(5t-1\right)\left(2t+3\right)$$
Rewrite the complete factored expression.
$$3\left(5t-1\right)\left(2t+3\right)$$
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 $ax^{2}+bx+c=0$.
$$30t^{2}+39t-9=0$$
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 $t=\frac{-39±51}{60}$ when $±$ is plus. Add $-39$ to $51$.
$$t=\frac{12}{60}$$
Reduce the fraction $\frac{12}{60}$ to lowest terms by extracting and canceling out $12$.
$$t=\frac{1}{5}$$
Now solve the equation $t=\frac{-39±51}{60}$ when $±$ is minus. Subtract $51$ from $-39$.
$$t=-\frac{90}{60}$$
Reduce the fraction $\frac{-90}{60}$ to lowest terms by extracting and canceling out $30$.
$$t=-\frac{3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1}{5}$ for $x_{1}$ and $-\frac{3}{2}$ for $x_{2}$.
Multiply $\frac{5t-1}{5}$ times $\frac{2t+3}{2}$ 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 $30$
$$x ^ 2 +\frac{13}{10}x -\frac{3}{10} = 0$$
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 $rs = C$
$$r + s = -\frac{13}{10} $$ $$ rs = -\frac{3}{10}$$
Two numbers $r$ and $s$ sum up to $-\frac{13}{10}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{13}{10} = -\frac{13}{20}$. 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 $u$.
