Factor the expression by grouping. First, the expression needs to be rewritten as $5x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-14$$ $$ab=5\left(-3\right)=-15$$
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 $-15$.
$$1,-15$$ $$3,-5$$
Calculate the sum for each pair.
$$1-15=-14$$ $$3-5=-2$$
The solution is the pair that gives sum $-14$.
$$a=-15$$ $$b=1$$
Rewrite $5x^{2}-14x-3$ as $\left(5x^{2}-15x\right)+\left(x-3\right)$.
$$\left(5x^{2}-15x\right)+\left(x-3\right)$$
Factor out $5x$ in $5x^{2}-15x$.
$$5x\left(x-3\right)+x-3$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(5x+1\right)$$
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 $ax^{2}+bx+c=0$.
$$5x^{2}-14x-3=0$$
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 $x=\frac{14±16}{10}$ when $±$ is plus. Add $14$ to $16$.
$$x=\frac{30}{10}$$
Divide $30$ by $10$.
$$x=3$$
Now solve the equation $x=\frac{14±16}{10}$ when $±$ is minus. Subtract $16$ from $14$.
$$x=-\frac{2}{10}$$
Reduce the fraction $\frac{-2}{10}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{1}{5}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $3$ for $x_{1}$ and $-\frac{1}{5}$ for $x_{2}$.
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 $5$
$$x ^ 2 -\frac{14}{5}x -\frac{3}{5} = 0$$
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 $rs = C$
$$r + s = \frac{14}{5} $$ $$ rs = -\frac{3}{5}$$
Two numbers $r$ and $s$ sum up to $\frac{14}{5}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{14}{5} = \frac{7}{5}$. 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 $u$.
$$r = \frac{7}{5} - u$$ $$s = \frac{7}{5} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -\frac{3}{5}$
