Math  /  Algebra

QuestionFactor completely: (x+6)6+(x+6)7(x+6)^{6}+(x+6)^{7}
Answer Attempt 1 out of 3

Studdy Solution

STEP 1

What is this asking? We need to rewrite this expression with the greatest common factor pulled out. Watch out! Don't forget to factor *completely*!

STEP 2

1. Find the Greatest Common Factor (GCF).
2. Factor out the GCF.
3. Simplify the remaining expression.

STEP 3

Let's look at our expression: (x+6)6+(x+6)7(x+6)^{6}+(x+6)^{7}.
We've got two terms here, and they both have a common factor of (x+6)(x+6) raised to some power.

STEP 4

Remember, the **greatest common factor** is the *largest* thing that divides *evenly* into both terms.
In this case, it's (x+6)(x+6) raised to the *smallest* exponent, which is **6**.
So, our **GCF** is (x+6)6(x+6)^6.

STEP 5

Now, let's **pull out** that (x+6)6(x+6)^6 from both terms!
Remember, when we factor something out, we're essentially *dividing* each term by the GCF.

STEP 6

So, we have (x+6)6(x+6)6(x+6)6+(x+6)6(x+6)7(x+6)6(x+6)^6 \cdot \frac{(x+6)^6}{(x+6)^6} + (x+6)^6 \cdot \frac{(x+6)^7}{(x+6)^6}.

STEP 7

Let's simplify those fractions!
Remember, anything divided by itself is **1**, so (x+6)6(x+6)6=1\frac{(x+6)^6}{(x+6)^6} = 1.

STEP 8

For the second fraction, (x+6)7(x+6)6\frac{(x+6)^7}{(x+6)^6}, we can use our exponent rules!
When we divide terms with the same base, we *subtract* the exponents.
So, 76=17-6=1, and we're left with (x+6)1(x+6)^1, which is just (x+6)(x+6).

STEP 9

Now, putting it all together, we have (x+6)6(1+(x+6))(x+6)^6(1 + (x+6)).

STEP 10

We can simplify the expression inside the parentheses: 1+(x+6)=x+71 + (x+6) = x + 7.

STEP 11

So, our **completely factored expression** is (x+6)6(x+7)(x+6)^6(x+7).
Awesome!

STEP 12

The completely factored form of (x+6)6+(x+6)7(x+6)^{6}+(x+6)^{7} is (x+6)6(x+7)(x+6)^6(x+7).

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord