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Chapter 6: Problem 73

Factor. If \(Q(x)=x^{6},\) use factoring to simplify $$ Q(a+h)-Q(a) $$

Short Answer

Expert verified
(a+h)^6 - a^6 = h((a+h)^5 + (a+h)^4a + (a+h)^3a^2 + (a+h)^2a^3 + (a+h)a^4 + a^5)

Step by step solution

01

Expand the expressions

Start by expanding both terms within the expression. Given that Q(x) = x^6, substitute x with (a+h) and a respectively. Q(a+h) = (a+h)^6 Q(a) = a^6
02

Write the difference

Substitute the expanded forms back into the expression Q(a+h) - Q(a): (a+h)^6 - a^6
03

Apply the difference of powers formula

Use the formula for the difference of two powers to factor the expression. The difference of two sixth powers can be factored as: a^6 - b^6 = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5) Use this in the context, where a = (a+h) and b = a. (a+h)^6 - a^6 = [h][(a+h)^5 + (a+h)^4a + (a+h)^3a^2 + (a+h)^2a^3 + (a+h)a^4 + a^5]
04

Simplify the result

The factored form of the expression is therefore: [h][(a+h)^5 + (a+h)^4a + (a+h)^3a^2 + (a+h)^2a^3 + (a+h)a^4 + a^5]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Powers
In algebra, the difference of powers is a valuable technique to simplify expressions. It refers to subtracting one power of a number or variable from another. For example, consider the general form for the difference of two fourth powers: a^4 - b^4In this case, the expression can be factored as: a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)The same principle applies to higher powers, just as in our problem. We see that:(a+h)^6 - a^6can be factored using the difference of two sixth powers equation, which is:a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)This formula is crucial for breaking down otherwise complex polynomials into simpler, more manageable expressions.
Polynomial Expansion
Polynomial expansion involves expanding an expression raised to a certain power into multiple terms. In the given problem, we need to expand:a^6 and (a+h)^6Using the binomial theorem, (a+h)^6 can be expanded as follows:a^6 + 6a^5h + 15a^4h^2 + 20a^3h^3 + 15a^2h^4 + 6ah^5 + h^6This allows us to see each term clearly and helps in further steps of factoring. The expansion simplifies into individual terms that contribute to the overall polynomial expression. Breaking down these expressions enables us to make use of known formulas like the difference of powers.
Simplification
Simplification is the process of reducing a mathematical expression to its simplest form. This often involves combining like terms and using algebraic identities and formulas. From our expanded and factored expression,a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)we substitute a with (a+h) and b with a, resulting in:(h)[(a+h)^5 + (a+h)^4a + (a+h)^3a^2 + (a+h)^2a^3 + (a+h)a^4 + a^5]This expression simplifies the difference of (a+h)^6 - a^6 by factoring out the common term h. The resulting simplified form is easier to understand and work with, demonstrating the effectiveness of combining expansion, factoring, and difference of powers techniques.

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Most popular questions from this chapter

To prepare for Section \(6.5,\) review the product and power rules for exponents and multiplication of polynomials (Sections 5.1 and 5.5 ). Multiply.[ 5.5]. $$ (x+1)(x+1)(x+1) $$

Factor completely. Remember to look first for a common factor. If a polynomial is prime, state this. $$ x^{2}+2 x y+y^{2}-9 $$

Factor completely. Remember to look first for a common factor. If a polynomial is prime, state this. $$ p^{2} q-25 q+3 p^{2}-75 $$

Solve. Round any irrational solutions to the nearest thousandth. $$ 3 x^{2}+x-1=0 $$

Factor completely. Remember to look first for a common factor. If a polynomial is prime, state this. $$ 49 a^{4}+100 $$
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