Solve 10+21x+2x^2

$$10 + 21 x + 2 x ^ { 2 }$$

Factor

$$\left(x+10\right)\left(2x+1\right)$$

Solution Steps

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

$$2x^{2}+21x+10$$

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

$$a+b=21 ab=2\times 10=20$$

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 20.

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

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 21.

$$a=1 b=20$$

Rewrite 2x^{2}+21x+10 as \left(2x^{2}+x\right)+\left(20x+10\right).

$$\left(2x^{2}+x\right)+\left(20x+10\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(2x+1\right)+10\left(2x+1\right)$$

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

$$\left(2x+1\right)\left(x+10\right)$$

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.

$$2x^{2}+21x+10=0$$

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.

$$x=\frac{-21±\sqrt{21^{2}-4\times 2\times 10}}{2\times 2}$$

Square 21.

$$x=\frac{-21±\sqrt{441-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-21±\sqrt{441-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$x=\frac{-21±\sqrt{441-80}}{2\times 2}$$

Add 441 to -80.

$$x=\frac{-21±\sqrt{361}}{2\times 2}$$

Take the square root of 361.

$$x=\frac{-21±19}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-21±19}{4}$$

Now solve the equation x=\frac{-21±19}{4} when ± is plus. Add -21 to 19.

$$x=-\frac{2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-21±19}{4} when ± is minus. Subtract 19 from -21.

$$x=-\frac{40}{4}$$

Divide -40 by 4.

$$x=-10$$

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 -10 for x_{2}.

$$2x^{2}+21x+10=2\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\left(-10\right)\right)$$

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

$$2x^{2}+21x+10=2\left(x+\frac{1}{2}\right)\left(x+10\right)$$

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

$$2x^{2}+21x+10=2\times \frac{2x+1}{2}\left(x+10\right)$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2x^{2}+21x+10=\left(2x+1\right)\left(x+10\right)$$

Evaluate

$$\left(x+10\right)\left(2x+1\right)$$