/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 (a)(i) Expand \((a+b)^{2}\) and ... [FREE SOLUTION] | 91Ó°ÊÓ

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(a)(i) Expand \((a+b)^{2}\) and \((a+b)^{3}\). (ii) Without doing any more work, write out the expanded forms of \((a-b)^{2}\) and \((a-b)^{3}\). (b) Factorise (i) \(x^{2}+2 x+1\) (ii) \(x^{4}-2 x^{2}+1\) (iii) \(x^{6}-3 x^{4}+3 x^{2}-1\). (c)(i) Expand \((a-b)(a+b)\). (ii) Use (c)(i) and (a)(i) to write down (with no extra work) the expanded form of $$ (a-b-c)(a+b+c) $$ and of $$ (a-b+c)(a+b-c) $$ (d) Factorise \(3 x^{2}+2 x-1\).

Short Answer

Expert verified
(a) (i): \((a+b)^2 = a^2 + 2ab + b^2\), \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\). (ii): \((a-b)^2 = a^2 - 2ab + b^2\), \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\). (b): (i) \((x+1)^2\), (ii) \((x^2-1)^2\), (iii) \((x^2-1)^3\). (c): (i) \(a^2-b^2\), (ii) \(a^2 -(b+c)^2\), \(a^2-(b-c)^2\). (d): \((3x-1)(x+1)\).

Step by step solution

01

Expand \((a+b)^2\)

Using the formula \((a+b)^2 = a^2 + 2ab + b^2\), expand the expression:\[(a+b)^2 = a^2 + 2ab + b^2\]
02

Expand \((a+b)^3\)

Using the formula \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\), expand the expression:\[(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\]
03

Use symmetry for \((a-b)^2\) and \((a-b)^3\)

For \((a-b)^2\), use the same terms but change the sign of the middle term:\[(a-b)^2 = a^2 - 2ab + b^2\]Similarly, for \((a-b)^3\), change sign in alternation:\[(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\]
04

Factorise \(x^2 + 2x + 1\)

Recognize as a perfect square trinomial:\[x^2 + 2x + 1 = (x+1)^2\]
05

Factorise \(x^4 - 2x^2 + 1\)

Notice that it is similar to a perfect square:\[x^4 - 2x^2 + 1 = (x^2 - 1)^2\]
06

Factorise \(x^6 - 3x^4 + 3x^2 - 1\)

Recognize it as \((x^2 - 1)^3\), using the pattern of cubic identities:\[x^6 - 3x^4 + 3x^2 - 1 = (x^2 - 1)^3\]
07

Expand \((a-b)(a+b)\)

Use the difference of squares formula:\[(a-b)(a+b) = a^2 - b^2\]
08

Analyze \((a-b-c)(a+b+c)\)

Recognize the similarity to \((a-b)(a+b)\). This expands using the difference-like structure:\[(a-b-c)(a+b+c) = a^2 - (b+c)^2\]Expanding further:\[a^2 - (b^2 + 2bc + c^2)\]
09

Analyze \((a-b+c)(a+b-c)\)

This also fits a difference-like structure:\[(a-b+c)(a+b-c) = a^2 - (b-c)^2\]Which becomes:\[a^2 - (b^2 - 2bc + c^2) = a^2 - b^2 + 2bc - c^2\]
10

Factorise \(3x^2 + 2x - 1\)

Look for two numbers that multiply to \(3 \times (-1) = -3\) and sum to \(2\). They are \(3\) and \(-1\):\[3x^2 + 3x - x - 1 = (3x^2 + 3x) - (x + 1)\]Factoring by grouping:\[3x(x + 1) - 1(x + 1) = (3x - 1)(x + 1)\]

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Key Concepts

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

Binomial Expansion
Binomial expansion is a fundamental concept in algebra that involves expanding expressions raised to a certain power. A binomial expression is typically in the form of \((a+b)^n\),where \(a\) and \(b\) are terms of the binomial, and \(n\) is the exponent. The expansion of such expressions can be handled systematically using the binomial theorem, which states that:\[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]Where \(\binom{n}{k}\) is a combination coefficient that determines how many ways \(k\) items can be chosen from \(n\). This formula enables us to expand expressions like \((a+b)^2\) and \((a+b)^3\) easily, giving us:
  • \((a+b)^2 = a^2 + 2ab + b^2\)
  • \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
For the expressions like \((a-b)^2\) or \((a-b)^3\), you can apply symmetry by changing the middle terms' signs, thus:
  • \((a-b)^2 = a^2 - 2ab + b^2\)
  • \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\)
This intuitive method makes working with binomial expansions more accessible.
Factoring Polynomials
Factoring polynomials involves simplifying a polynomial expression into a product of simpler polynomials. This technique helps solve equations and simplify expressions. Always look for common factors first. For example, consider a quadratic polynomial like \(x^2 + 2x + 1\). This is a perfect square trinomial that can be factored into:\[(x + 1)^2\]Similarly, for a polynomial like \(x^4 - 2x^2 + 1\), recognizing it as a perfect square gives:\[(x^2 - 1)^2\]More complex polynomials, such as \(x^6 - 3x^4 + 3x^2 - 1\), can often be broken down using your knowledge of patterns. Recognizing it as related to cubic patterns, it can be factored as:\[(x^2 - 1)^3\]Looking for recognizable forms and identities is a strategic approach in factoring polynomials.
Perfect Squares
A perfect square trinomial is a specific kind of trinomial that is the result of squaring a binomial. They are important because they simplify the factoring process significantly. When you see a trinomial of the form \(x^2 + 2ax + a^2\), it can immediately be factored into:\[(x + a)^2\]These trinomials are easily identified because the square of the first term and the last term, with the middle term being twice the product of their roots. For instance, \(x^2 + 2x + 1\) can be observed as:
  • First term: \(x^2 = (x)^2\)
  • Middle term: \(2x = 2 \times x \times 1\)
  • Last term: \(1 = (1)^2\)
Perfect squares make algebraic expression manipulation more straightforward and are foundational in many algebraic solutions.
Difference of Squares
The difference of squares is a special algebraic identity that allows certain polynomial expressions to be easily factored. This identity states that any expression in the form:\[a^2 - b^2\]can be factored into:\[(a-b)(a+b)\]This property is incredibly useful when dealing with terms that fit this pattern, as it simplifies the factorization process considerably. For example, \((a-b)(a+b) = a^2 - b^2\).Consider another application such as factorizing expressions like \((a-b-c)(a+b+c)\).This can be reinterpreted as a difference-type arrangement similar to what is seen in:\[a^2 - (b+c)^2\]Utilizing such patterns makes the analysis and simplification process intuitive and systematic.

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Most popular questions from this chapter

Write out the first 12 or so powers of 4 : $$ 4,16,64,256,1024,4096,16384,65536, \ldots $$ Now create two sequences: the sequence of final digits: \(4,6,4,6,4,6, \ldots\) the sequence of leading digits: \(4,1,6,2,1,4,1,6, \ldots\) Both sequences seem to consist of a single "block", which repeats over and over for ever. (a) How long is the apparent repeating block for the first sequence? How long is the apparent repeating block for the second sequence? (b) It may not be immediately clear whether either of these sequences really repeats forever. Nor may it be clear whether the two sequences are alike, or whether one is quite different from the other. Can you give a simple proof that one of these sequences recurs, that is, repeats forever? (c) Can you explain why the other sequence seems to recur, and decide whether it really does recur forever?

(a) In the "24 game" you are given four numbers. Your job is to use each number once, and to combine the four numbers using any three of the four basic arithmetical operations - using the same operation more than once if you wish, and as many brackets as you like (but never concatenating different numbers, such as "3" and "4" to make " \(34 "\) ). If the given numbers are \(3,3,4,4,\) then one immediately sees \(3 \times 4+3 \times 4=24 .\) With 3,3,5 , 5 it may take a little longer, but is still fairly straightforward. However, you may find it more challenging to make 24 in this way: (i) using the four numbers 3,3,6,6 (ii) using the four numbers 3,3,7,7 (iii) using the four numbers 3,3,8,8 . (b) Suppose we restrict the numbers to be used each time to "four \(4 \mathrm{~s}\) " \((4,4,4,4),\) and change the goal from "make \(24 ",\) to "make each answer from \(0-10\) using exactly four \(4 \mathrm{~s} "\). (i) Which of the numbers \(0-10\) cannot be made? (ii) What if one is allowed to use squaring and square roots as well as the four basic operations? What is the first inaccessible integer? Calculating by turning the handle deterministically (as with addition, or multiplication, or multiplying out brackets, or differentiating) is a valuable skill. But such direct procedures are usually only the beginning. Using mathematics and solving problems generally depend on the corresponding inverse procedures \(-\) where a certain amount of juggling and insight is needed in order to work backwards (as with subtraction, or division, or factorisation, or integration). For example, in applications of calculus, the main challenge is to solve differential equations (an inverse problem) rather than to differentiate known functions. Problem 14 captures the spirit of this idea in the simplest possible context of arithmetic: the required answer is given, and we have to find how (or whether) that answer can be generated. We will meet more interesting examples of this kind throughout the rest of the collection.

The 4 by 4 "multiplication table" below is completely familiar. \(\begin{array}{rrrr}1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 4 & 8 & 12 & 16\end{array}\) What is the total of all the numbers in the 4 by 4 square? How should one write this answer in a way that makes the total obvious?

Let \(A B C\) be a triangle. We use the standard labelling convention, whereby the side \(B C\) opposite \(A\) has length \(a\), the side \(C A\) opposite \(B\) has length \(b,\) and the side \(A B\) opposite \(C\) has length \(c\). Prove that, if \(c^{2}=a^{2}+b^{2},\) then \(\angle B C A\) is a right angle.

(a) Compute by mental arithmetic (using pencil only to record results), then learn by heart: (i) the squares of positive integers: first up to \(12^{2}\); then to \(31^{2}\) (ii) the cubes of positive integers up to \(11^{3}\) (iii) the powers of 2 up to \(2^{10}\). (b) How many squares are there: (i) \(<1000 ?\) (ii) \(<10000 ?\) (iii) \(<100000 ?\) (c) How many cubes are there: (i) (ii) (iii) \(<1000000 ?\) (d) (i) Which powers of 2 are squares? (ii) Which powers of 2 are cubes? (e) Find the smallest square greater than 1 that is also a cube. Find the next smallest. Evaluating powers, and the associated index laws, constitute an example of a direct operation. For each direct operation, we need to think carefully about the corresponding inverse operation - here "extracting roots". In particular, we need to be clear about the distinction between the fact that the equation \(x^{2}=4\) has two different solutions, while \(\sqrt{4}\) has just one value (namely 2).

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