How do you multiply (3x2+2x+4)(2x+1)?

Question

Zayden Sims · Accepted Answer

Explanation:
You have to take each number in the first bracket and multiply it by the second one.
First lets

d o 3 x^{2}

(3 x^{2}) \cdot (2 x)

and

(3 x^{2}) \cdot 1

This will give

6 x^{3}

and

3 x^{2}

Then 2x
2x*2x and 2x*1
This will give

4 x^{2}

and 2x
Then 4
4*2x and 4*1
This will give 8x and 4

ribejat1f3 · Accepted Answer

Explanation:
Do the first term in the first bracket multiplied by the first term in the second bracket. Do the same with the first term in the first bracket $(3 x^{2})$ and the second term of the second bracket.
$(3 x^{2}) \times (2 x) = 6 x^{3}$
$(3 x^{2}) \times (1) = 3 x^{2}$
For powers, use the rule:
$a^{m} \times a^{n} = a^{m} + n$ (where x is actually $x^{1}$ from example above which simply calculates to x)
Then multiply any real numbers as you normally would $(3 \times 2)$
For the next step, multiply the second term of the first bracket to the first of the second bracket:
$(2 x) \times (2 x) = 4 x^{2}$
$(2 x) \times (1) = 2 x$
Proceed to the final term in the first bracket and follow the same steps:
$(4) \times (2 x) = 8 x$
$(4) \times (1) = 4$
Simplify by adding all like terms together:
$= 6 x^{3} + 3 x^{2} + 4 x^{2} + 2 x + 8 x + 4$
$= 6 x^{3} + 7 x^{2} + 10 x + 4$

How do you multiply (3x^{2}+2x+4)(2x+1)?

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