Simplify 3b^3 × b^7 × b^8 × b^10 + 30b^14: Exponent Practice

Question

3b3×b7×b8×b10+30b14= 3b^3\times b^7\times b^8\times b^{10}+30b^{14}=

Simplify the above expression as much as possible.

Video Solution

Solution Steps

00:00 Simply
00:03 Let's solve the product of numbers first
00:10 When multiplying powers with equal bases
00:13 The power of the result equals the sum of powers
00:17 We'll use this formula in our exercise and add the powers
00:39 Let's factor out the common term from the parentheses
00:51 And this is the solution to the question

Step-by-Step Solution

Let's first deal with the first term which is the multiplication term:

3b3b7b8b10 3b^3\cdot b^7\cdot b^8\cdot b^{10}

We'll handle separately the numbers and algebraic expressions (i.e. - the letters), noting that all algebraic multiplication terms have the same base, therefore we'll use the power law for multiplication between terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n}

Note that this law applies to any number of terms in multiplication and not just for two, for example for multiplication of three terms with identical base we get:

amanak=am+nak=am+n+k a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}

When we used the above power law twice, we can also perform the same calculation for four terms in multiplication five etc...

From here on we will no longer indicate the multiplication sign, but use the conventional writing form where placing terms next to each other means multiplication.

Let's return to the problem and apply the above power law for the multiplication term above:

3b3b7b8b10=3b3+7+8+10=3b28 3b^3b^7b^8b^{10}=3b^{3+7+8+10}=3b^{28}

Where we handled the numbers and letters separately.

Let's return to the original question and substitute the multiplication term with the result we got in the last step:

3b3b7b8b10+30b14=3b28+30b14 3b^3b^7b^8b^{10}+30b^{14}=3b^{28}+30b^{14}

Note that we can further simplify the expression by using factorization by taking out the greatest common factor, for the numbers - we can take out of the parentheses the greatest common factor of 30 and 3, which is 3, and for the letters, the greatest common factor we can take out is:

b14 b^{14}

Therefore the factorization we get is:

3b28+30b14=3b14(b14+10) 3b^{28}+30b^{14}=3b^{14}(b^{14}+10)

Where we used the power law for multiplication between terms with identical bases to know that:

b28=b14b14 b^{28}=b^{14}\cdot b^{14}

We got the most simplified and factored expression for the above problem:

3b14(b14+10) 3b^{14}(b^{14}+10)

Therefore the most correct answer is B

Small note:

You can always expand the parentheses after factorization to verify that the factorization was done correctly.

Answer

3b14(b14+10) 3b^{14}(b^{14}+10)