HSC Maths Extension 1 Notes

Assumed Knowledge Notes. The following very brief notes cover a few of the basic things you should know from Extension 1 Mathematics before attempting the Assumed Knowledge Quiz. For further assistance, you should consult a high school text book. You may also find the worksheets at http://www.maths.unsw.edu.au/currentstudents/revision-worksheets helpful. Polynomials. 1. Basics: A polynomial is an expression of the form p(x) = an xn + an−1 xn−1 + ... + a1 x + a0 where the co-efficients ai are real numbers. e.g. −5x3 + 2x + 3, 72 x7 +

√

1

2x3 . Note that x3 + 2x 2 − 1 is NOT a polynomial.

We say that n is the degree of the polynomial p(x), and write deg(p(x)) = n. If the polynomial is non-zero constant then it has degree 0. The zero polynomial does not have a degree. an is called the leading co-efficient and a0 is called the constant term. If an = 1 we say that the polynomial is monic. e.g. p(x) = x7 + 3x3 + 1 is a monic polynomial of degree 7. In this section we are going to see how to factor polynomials (if possible) and use this to solve polynomial equations. 2. Operations on Polynomials: We can add and subtract polynomials in the obvious way, for example (3x7 − 2x5 + x4 − 9x − 10) + (2x5 + 7x4 − 2x3 + 12) = 3x7 + 8x4 − 2x3 − 9x + 2. We multiply polynomials using long multiplication, e.g. (x3 − 2x2 + 5x − 1) × (x − 2) We simply expand the brackets to obtain x(x3 − 2x2 + 5x − 1) − 2(x3 − 2x2 + 5x − 1) = x4 − 4x3 + 9x2 − 11x + 2. Polynomials can also be divided using the long division algorithm from primary school. Ex: Divide x4 − 3x3 + 2x − 4 by (x − 1).

1

x−1

x3 − 2x2 − 2x )x4 − 3x3 + 0x2 + 2x − 4 x4 − x3 −2x3 + 0x2 −2x3 + 2x2 −2x2 + 2x −2x2 + 2x 0x − 4

The quotient is x3 − 2x2 − 2x and the remainder is −4. Thus we can write x4 − 3x3 + 2x − 4 = (x − 1)(x3 − 2x2 − 2x) − 4. We can do this for any pair of polynomials, in fact: Given polynomials a(x), b(x), we can find polynomials q(x) and r(x) such that a(x) = q(x)b(x) + r(x) where the degree of the remainder r(x) is less than the degree of the divisor b(x), or r(x) = 0. Ex: Divide x5 − 2x3 + 1 by x3 + 2. We do the division to obtain x5 − 2x3 + 1 = (x2 − 2)(x3 + 2) + (5 − 2x2 ) and the degree of the remainder is 2 which is less than the degree of the divisor, 3. 3. The Remainder and Factor Theorems: If we divide by a linear factor (x − a) then the remainder can be obtained without actually doing the division. Theorem: (The Remainder Theorem) If the polynomial p(x) is divided by x − a then the remainder is simply p(a). Proof: We write p(x) = (x − a)q(x) + r where r is a constant since it has degree smaller than 1, i.e. degree 0. Put x = a in both sides and the result follows. Ex: Find the remainder when p(x) = x3 − 2x2 + 5x − 1 is divided by x − 1. p(1) = 3 so the remainder is 3. Ex: Find the remainder when p(x) = x3 + 6x2 − 13x − 42 is divided by x + 2. p(−2) = −8+24+26−42 = 0 so the remainder is 0. This means that x+2 is a factor of p(x). If one polynomial q(x) exactly divides another polynomial p(x) we say that q is a factor of p. In particular, a linear polynomial x − a is a factor of a polynomial p(x) if the remainder 2

p(a) is zero. We can exploit this idea to factorise polynomials. Theorem: (The Factor Theorem) If p(α) = 0 then x − α is a factor of p(x). Ex: Factorise p(x) = x3 − 8x2 + 5x + 14. p(1) 6= 0, p(−1) = 0 so (x + 1) is a factor. Also p(2) = 0 so x − 2 is a factor. We can divide p(x) by (x + 1)(x − 2) to obtain p(x) = (x − 2)(x + 1)(x − 7). Ex: Solve x3 − 8x2 + 5x + 14 = 0. In factored form this says (x − 2)(x + 1)(x − 7) = 0 and so x = 2, −1, 7. In practice, if the constant term is a, we try all the (positive and negative) factors of a. Ex: Solve p(x) = x3 + 4x2 − 7x − 10 = 0. p(1) 6= 0, p(−1) = 0 so p(x) = (x + 1)(x2 + 3x − 10) = (x + 1)(x + 5)(x − 2) and hence the roots are −1, −5, 2. The Binomial Theorem. Please read the Sheet on The Binomial Theorem at http://www.maths.unsw.edu.au/currentstudents/revision-worksheets

The Factorial Function and n Cr Notation: The notation n! means we multiply n(n − 1)(n − 2)...4.3.2.1. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. This in fact counts the number of ways that 5 people can stand in a straight line. We read n! as ‘n factorial’ and this function is built into your calculator. Note that we define 0! to be 1. n r

n

We also use the notation Cr also written as n

Cr =

!

to mean

n! r!(n − r)!

Ex. Find 8 C3 , 5 C0 . From the calculator (or by hand) 8 C3 = 56 and 5 C0 = 1.

3

The numbers in Pascal’s triangle are given by the n Cr formula. In fact, we can write (1 + x) = 1 C0 + 1 C1 x (1 + x)2 = 2 C0 + 2 C1 x + 2 C2 x2 (1 + x)3 = 3 C0 + 3 C1 x + 3 C2 x2 + 3 C3 x3 (1 + x)4 = 4 C0 + 4 C1 x + 4 C2 x2 + 4 C3 x3 + 4 C4 x4 and so on. The general form is (1 + x)n =

n

C0 +

n

C1 x +

n

C2 x2 + ... +

n

Cn xn

and more generally again, we have (a + b)n =

n

C0 an +

n

C1 an−1 b +

n

C2 an−2 b2 + ... +

n

Cn bn

Generally, when expanding out we use Pascal’s triangle rather than the n Cr formula, but the formula is very useful in doing other more theoretical problems. The general term in the expansion of (a + b)n is ! n n n−r r an−r br . Cr a b = r Ex: Find the co-efficient of x8 and the constant term in the expansion of (2x3 − x1 )12 . 12 r

!

!

12 (2x ) × 212−r × x36−4r × (−1)r . Now The general term is = r ! 12 36 − 4r = 8 when r = 7 and so the coefficient of x8 is 25 × (−1)7 = −25344. 7 3 12−r

(− x1 )r

The constant term will appear when 36 − 4r = 0 and so r = 9. Hence the constant term is ! 12 23 × (−1)9 = −1760. 9 Integration by Substitution. Many integrals can be performed by making a change of variable. Z

Ex. Find I =

2

xex dx, using u = x2 .

Put u = x2 , then then becomes

du dx

= 2x. Hence we can symbolically replace x dx by I=

1Z u 1 1 2 e du = eu + C = ex + C. 2 2 2

Harder Inequalities. Please read the Sheet on Polynomial Inequalities at 4

du . 2

The integral

http://www.maths.unsw.edu.au/currentstudents/revision-worksheets Ex. Solve

x−1 x+1

≤ 5.

To solve this we multiply top and bottom by the square of (x + 1) which will not affect the inequality sign, since this quantity is positive (provided x 6= −1). Thus x−1 x−1 ≤5⇒ × (x + 1)2 ≤ 5 × (x + 1)2 ⇒ (x − 1)(x + 1) ≤ 5(x + 1)2 . x+1 x+1 Now move the terms to the right hand side and factorise, giving (x + 1)[(5(x + 1) − (x − 1)] ≥ 0 ⇒ (x + 1)(4x + 6) ≥ 0. Thus the solution is x > −1 or x ≤ − 32 . This last step is done by sketching the graph of y = (x + 1)(4x + 6), as shown in the sheet on Polynomial Inequalities. Note that you CANNOT multiply by (x + 1) as a first step since we do not know if this is positive or negative. Further Trigonometry Please read the sheet on Trigonometric Identities at http://www.maths.unsw.edu.au/currentstudents/revision-worksheets Ex. Convert sin x − cos x into the form R sin(x + α). Expanding R sin(x + α) we have R sin x cos α + R cos x sin α = sin x − cos x. Now equate the cooefficients of sin x and cos x to obtain R sin α = −1.

R cos α = 1

Squaring the adding these equations gives R2 = 2 so we may take R = equations gives tan α = −1 and so we may take α = − π4 . Hence sin x − cos x =

√

2 sin(x − π4 ).

Note that this gives us the amplitude of the wave sin x − cos x to be

5

√

2.

√

2. Dividing the