Solve 40x^2+14x+1

$$40 { x }^{ 2 } +14x+1$$

Factor

$$\left(4x+1\right)\left(10x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 40x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=14 ab=40\times 1=40$$

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 40.

$$1,40 2,20 4,10 5,8$$

Calculate the sum for each pair.

$$1+40=41 2+20=22 4+10=14 5+8=13$$

The solution is the pair that gives sum 14.

$$a=4 b=10$$

Rewrite 40x^{2}+14x+1 as \left(40x^{2}+4x\right)+\left(10x+1\right).

$$\left(40x^{2}+4x\right)+\left(10x+1\right)$$

Factor out 4x in 40x^{2}+4x.

$$4x\left(10x+1\right)+10x+1$$

Factor out common term 10x+1 by using distributive property.

$$\left(10x+1\right)\left(4x+1\right)$$

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.

$$40x^{2}+14x+1=0$$

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.

$$x=\frac{-14±\sqrt{14^{2}-4\times 40}}{2\times 40}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 40}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-14±\sqrt{196-160}}{2\times 40}$$

Add 196 to -160.

$$x=\frac{-14±\sqrt{36}}{2\times 40}$$

Take the square root of 36.

$$x=\frac{-14±6}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{-14±6}{80}$$

Now solve the equation x=\frac{-14±6}{80} when ± is plus. Add -14 to 6.

$$x=-\frac{8}{80}$$

Reduce the fraction \frac{-8}{80} to lowest terms by extracting and canceling out 8.

$$x=-\frac{1}{10}$$

Now solve the equation x=\frac{-14±6}{80} when ± is minus. Subtract 6 from -14.

$$x=-\frac{20}{80}$$

Reduce the fraction \frac{-20}{80} to lowest terms by extracting and canceling out 20.

$$x=-\frac{1}{4}$$

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{1}{4} for x_{2}.

$$40x^{2}+14x+1=40\left(x-\left(-\frac{1}{10}\right)\right)\left(x-\left(-\frac{1}{4}\right)\right)$$

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

$$40x^{2}+14x+1=40\left(x+\frac{1}{10}\right)\left(x+\frac{1}{4}\right)$$

Add \frac{1}{10} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$40x^{2}+14x+1=40\times \frac{10x+1}{10}\left(x+\frac{1}{4}\right)$$

Add \frac{1}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$40x^{2}+14x+1=40\times \frac{10x+1}{10}\times \frac{4x+1}{4}$$

Multiply \frac{10x+1}{10} times \frac{4x+1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$40x^{2}+14x+1=40\times \frac{\left(10x+1\right)\left(4x+1\right)}{10\times 4}$$

Multiply 10 times 4.

$$40x^{2}+14x+1=40\times \frac{\left(10x+1\right)\left(4x+1\right)}{40}$$

Cancel out 40, the greatest common factor in 40 and 40.

$$40x^{2}+14x+1=\left(10x+1\right)\left(4x+1\right)$$

Evaluate

$$\left(4x+1\right)\left(10x+1\right)$$