Solve 3+11x+10x^2

$$3+11x+10 { x }^{ 2 }$$

Factor

$$\left(2x+1\right)\left(5x+3\right)$$

Solution Steps

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

$$10x^{2}+11x+3$$

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=11 ab=10\times 3=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=5 b=6$$

Rewrite 10x^{2}+11x+3 as \left(10x^{2}+5x\right)+\left(6x+3\right).

$$\left(10x^{2}+5x\right)+\left(6x+3\right)$$

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

$$5x\left(2x+1\right)+3\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(5x+3\right)$$

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.

$$10x^{2}+11x+3=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\times 10\times 3}}{2\times 10}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 10\times 3}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-11±\sqrt{121-40\times 3}}{2\times 10}$$

Multiply -40 times 3.

$$x=\frac{-11±\sqrt{121-120}}{2\times 10}$$

Add 121 to -120.

$$x=\frac{-11±\sqrt{1}}{2\times 10}$$

Take the square root of 1.

$$x=\frac{-11±1}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-11±1}{20}$$

Now solve the equation x=\frac{-11±1}{20} when ± is plus. Add -11 to 1.

$$x=-\frac{10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{-11±1}{20} when ± is minus. Subtract 1 from -11.

$$x=-\frac{12}{20}$$

Reduce the fraction \frac{-12}{20} to lowest terms by extracting and canceling out 4.

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

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

$$10x^{2}+11x+3=10\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\left(-\frac{3}{5}\right)\right)$$

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

$$10x^{2}+11x+3=10\left(x+\frac{1}{2}\right)\left(x+\frac{3}{5}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+11x+3=10\times \frac{2x+1}{2}\left(x+\frac{3}{5}\right)$$

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

$$10x^{2}+11x+3=10\times \frac{2x+1}{2}\times \frac{5x+3}{5}$$

Multiply \frac{2x+1}{2} times \frac{5x+3}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+11x+3=10\times \frac{\left(2x+1\right)\left(5x+3\right)}{2\times 5}$$

Multiply 2 times 5.

$$10x^{2}+11x+3=10\times \frac{\left(2x+1\right)\left(5x+3\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10x^{2}+11x+3=\left(2x+1\right)\left(5x+3\right)$$

Evaluate

$$\left(2x+1\right)\left(5x+3\right)$$