Solve 50r^2-3+25r

$$50 r ^ { 2 } - 3 + 25 r$$

Factor

$$\left(10r-1\right)\left(5r+3\right)$$

Solution Steps

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

$$50r^{2}+25r-3$$

Factor the expression by grouping. First, the expression needs to be rewritten as 50r^{2}+ar+br-3. To find a and b, set up a system to be solved.

$$a+b=25 ab=50\left(-3\right)=-150$$

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 -150.

$$-1,150 -2,75 -3,50 -5,30 -6,25 -10,15$$

Calculate the sum for each pair.

$$-1+150=149 -2+75=73 -3+50=47 -5+30=25 -6+25=19 -10+15=5$$

The solution is the pair that gives sum 25.

$$a=-5 b=30$$

Rewrite 50r^{2}+25r-3 as \left(50r^{2}-5r\right)+\left(30r-3\right).

$$\left(50r^{2}-5r\right)+\left(30r-3\right)$$

Factor out 5r in the first and 3 in the second group.

$$5r\left(10r-1\right)+3\left(10r-1\right)$$

Factor out common term 10r-1 by using distributive property.

$$\left(10r-1\right)\left(5r+3\right)$$

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.

$$50r^{2}+25r-3=0$$

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.

$$r=\frac{-25±\sqrt{25^{2}-4\times 50\left(-3\right)}}{2\times 50}$$

Square 25.

$$r=\frac{-25±\sqrt{625-4\times 50\left(-3\right)}}{2\times 50}$$

Multiply -4 times 50.

$$r=\frac{-25±\sqrt{625-200\left(-3\right)}}{2\times 50}$$

Multiply -200 times -3.

$$r=\frac{-25±\sqrt{625+600}}{2\times 50}$$

Add 625 to 600.

$$r=\frac{-25±\sqrt{1225}}{2\times 50}$$

Take the square root of 1225.

$$r=\frac{-25±35}{2\times 50}$$

Multiply 2 times 50.

$$r=\frac{-25±35}{100}$$

Now solve the equation r=\frac{-25±35}{100} when ± is plus. Add -25 to 35.

$$r=\frac{10}{100}$$

Reduce the fraction \frac{10}{100} to lowest terms by extracting and canceling out 10.

$$r=\frac{1}{10}$$

Now solve the equation r=\frac{-25±35}{100} when ± is minus. Subtract 35 from -25.

$$r=-\frac{60}{100}$$

Reduce the fraction \frac{-60}{100} to lowest terms by extracting and canceling out 20.

$$r=-\frac{3}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{10} for x_{1} and -\frac{3}{5} for x_{2}.

$$50r^{2}+25r-3=50\left(r-\frac{1}{10}\right)\left(r-\left(-\frac{3}{5}\right)\right)$$

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

$$50r^{2}+25r-3=50\left(r-\frac{1}{10}\right)\left(r+\frac{3}{5}\right)$$

Subtract \frac{1}{10} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$50r^{2}+25r-3=50\times \frac{10r-1}{10}\left(r+\frac{3}{5}\right)$$

Add \frac{3}{5} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$50r^{2}+25r-3=50\times \frac{10r-1}{10}\times \frac{5r+3}{5}$$

Multiply \frac{10r-1}{10} times \frac{5r+3}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$50r^{2}+25r-3=50\times \frac{\left(10r-1\right)\left(5r+3\right)}{10\times 5}$$

Multiply 10 times 5.

$$50r^{2}+25r-3=50\times \frac{\left(10r-1\right)\left(5r+3\right)}{50}$$

Cancel out 50, the greatest common factor in 50 and 50.

$$50r^{2}+25r-3=\left(10r-1\right)\left(5r+3\right)$$

Evaluate

$$\left(10r-1\right)\left(5r+3\right)$$