Solve 25x^2+145x+100

$$25 { x }^{ 2 } +145x+100$$

Factor

$$5\left(x+5\right)\left(5x+4\right)$$

Solution Steps

Factor out 5.

$$5\left(5x^{2}+29x+20\right)$$

Consider 5x^{2}+29x+20. Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=29 ab=5\times 20=100$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 100.

$$1,100 2,50 4,25 5,20 10,10$$

Calculate the sum for each pair.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

The solution is the pair that gives sum 29.

$$a=4 b=25$$

Rewrite 5x^{2}+29x+20 as \left(5x^{2}+4x\right)+\left(25x+20\right).

$$\left(5x^{2}+4x\right)+\left(25x+20\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(5x+4\right)+5\left(5x+4\right)$$

Factor out common term 5x+4 by using distributive property.

$$\left(5x+4\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$5\left(5x+4\right)\left(x+5\right)$$

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.

$$25x^{2}+145x+100=0$$

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.

$$x=\frac{-145±\sqrt{145^{2}-4\times 25\times 100}}{2\times 25}$$

Square 145.

$$x=\frac{-145±\sqrt{21025-4\times 25\times 100}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-145±\sqrt{21025-100\times 100}}{2\times 25}$$

Multiply -100 times 100.

$$x=\frac{-145±\sqrt{21025-10000}}{2\times 25}$$

Add 21025 to -10000.

$$x=\frac{-145±\sqrt{11025}}{2\times 25}$$

Take the square root of 11025.

$$x=\frac{-145±105}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{-145±105}{50}$$

Now solve the equation x=\frac{-145±105}{50} when ± is plus. Add -145 to 105.

$$x=-\frac{40}{50}$$

Reduce the fraction \frac{-40}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{4}{5}$$

Now solve the equation x=\frac{-145±105}{50} when ± is minus. Subtract 105 from -145.

$$x=-\frac{250}{50}$$

Divide -250 by 50.

$$x=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{4}{5} for x_{1} and -5 for x_{2}.

$$25x^{2}+145x+100=25\left(x-\left(-\frac{4}{5}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$25x^{2}+145x+100=25\left(x+\frac{4}{5}\right)\left(x+5\right)$$

Add \frac{4}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25x^{2}+145x+100=25\times \frac{5x+4}{5}\left(x+5\right)$$

Cancel out 5, the greatest common factor in 25 and 5.

$$25x^{2}+145x+100=5\left(5x+4\right)\left(x+5\right)$$

Evaluate

$$5\left(x+5\right)\left(5x+4\right)$$