Factor the expression by grouping. First, the expression needs to be rewritten as $25x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=25\left(-2\right)=-50$$
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 $-50$.
$$-1,50$$ $$-2,25$$ $$-5,10$$
Calculate the sum for each pair.
$$-1+50=49$$ $$-2+25=23$$ $$-5+10=5$$
The solution is the pair that gives sum $5$.
$$a=-5$$ $$b=10$$
Rewrite $25x^{2}+5x-2$ as $\left(25x^{2}-5x\right)+\left(10x-2\right)$.
$$\left(25x^{2}-5x\right)+\left(10x-2\right)$$
Factor out $5x$ in the first and $2$ in the second group.
$$5x\left(5x-1\right)+2\left(5x-1\right)$$
Factor out common term $5x-1$ by using distributive property.
$$\left(5x-1\right)\left(5x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$25x^{2}+5x-2=0$$
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 $x=\frac{-5±15}{50}$ when $±$ is plus. Add $-5$ to $15$.
$$x=\frac{10}{50}$$
Reduce the fraction $\frac{10}{50}$ to lowest terms by extracting and canceling out $10$.
$$x=\frac{1}{5}$$
Now solve the equation $x=\frac{-5±15}{50}$ when $±$ is minus. Subtract $15$ from $-5$.
$$x=-\frac{20}{50}$$
Reduce the fraction $\frac{-20}{50}$ to lowest terms by extracting and canceling out $10$.
$$x=-\frac{2}{5}$$
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 $-\frac{2}{5}$ for $x_{2}$.
Multiply $\frac{5x-1}{5}$ times $\frac{5x+2}{5}$ 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 $25$
$$x ^ 2 +\frac{1}{5}x -\frac{2}{25} = 0$$
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 $rs = C$
$$r + s = -\frac{1}{5} $$ $$ rs = -\frac{2}{25}$$
Two numbers $r$ and $s$ sum up to $-\frac{1}{5}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{1}{5} = -\frac{1}{10}$. 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 $u$.
