IB Mathematics AA HL Flashcards- Polynomials

↳ Degree of a polynomial is its largest exponent: n
↳ Quadratics/ have degree of 2 (ax² + bx + c). Dealt with in 2.7
↳ CUBICS/ ODD, ∀ x∈ℝ √-x = -√x, ≤ 3 turns QUARTIC / EVEN, ∀ x∈ℝ √-x = √x, ≤ 3 turns

FEATURES // Smooth curves // Continuous (no gaps)
// It rises (p(x) → ∞) or falls (p(x) → -∞) without bound as x→±∞
// Domain is x∈ℝ // A polynomial with degree n has at most n-1 turns.

REMAINDER THEOREM // Just as you do 65÷7 and get an answer such as 9 rem 2,
// we can do the same for dividing a polynomial by (x-c). NOT in F.B.
// If a polynomial, p(x), is divided by (x-c) the remainder is p(c)

[q]
DIVISION // We can test potential factors, or find remainders/quotients, by doing
// synthetic (or long division. (Below, we have (2x³+5x²+6x-3)÷(x+2) )

[a]

FACTOR
THEOREM // (x-c) is a factor of polynomial P(x) if & only if P(c) = 0 NOT in F.B.
// This is really just an extension of the remainder theorem, as it mates
// sense that if there is no remainder, we have a factor.

E.G. // a) Show that (x-4) is a factor of f(x) = x³+5x²-36:
↳ f(4) = (4)³+5(4)²-36 = 64+20-36 = 0, hence, it is a factor

[q]

// b) Using this knowledge, with synthetic div., we can solve harder problems:

E.G. // Given that x=-2 & x=3 are zeros of the polynomial function:
// h(x) = x⁴+2x³-13x²-14x+24, find the other 2 zeros of h(x):

(Two methods are shown. One is a longer, manual method, and the other is a quicker method using synthetic division.)

[a]

↳ We can do this manually:
h(x) = x⁴+2x³-13x²-14x+24 = (x+2)(x-3)(ax²+bx+c)
= (x²+…)(ax²+bx+c)

[q]

Now, we compare coefficients:
Constant term: -16 = (c)(+2)(-3) ⇒ c= +8/3
x³ term: (2ax³)+(bx³) = 2x³ ⇒ (2a+b)x³ = 2x³
⇒ -6a+3b = 6 & 2a+b = 2
⇒ b=0

… h(x) = (x+2)(x-3)(x²+8/3)
Solving x²+8/3 = 0, gives x = ±2i

OR Using synthetic division may
be quicker and easier:

[a]
QUADRATIC // When we have a quadratic equation in the form
// y = ax²+bx+c, we know the following:

SUM OF ROOTS = -b/a PRODUCT OF ROOTS = c/a

[q]

PROOF // ax² + bx + c = 0 ⇒ x² + (b/a)x + c/a = 0
// x² + (b/a)x + c/a = (x-α)(x-β) = 0
// x² + (b/a)x + c/a = x² – (α+β)x + αβ = 0
// ↳ Equating coefficients: α+β = -b/a , αβ = c/a

E.G. 5// Find the sum & product of roots of f(x) = 3x²-6x+15:
↳ SUM: -b/a = -(-6)/3 = 2 PRODUCT: c/a = 15/3 = 5

[a]

POLYNOMIAL ROOTS // A natural extension of the above gives us:
For p(x) = anxn+an-1xn-1+…+a1x+a0, SUM OF ROOTS = -an-1/an PROD. OF ROOTS = (-1)na0/an

E.G. 6// If x³-3x²+kx+75 = 0 has roots α, β, γ, find k:
↳ SUM: α+β+γ = -(-3)/1 = 3 PRODUCT: α·β·γ = -75/1 = -75
// Roots: 5, -5, 3
// -α²·γ = -25
// -k=25
// k = -25

… x³-3x²+kx+75 = (x-5)(x+5)(x-3) ⇒ x³-3x²-25x+75 ⇒ k=-25

[q]

CONJUGATE ROOTS //
// ↳ A polynomial of degree n>0 has exactly n roots.
// ↳ If the polynomial has an odd degree, as complex roots come in pairs,
// then it must have at least one real root.

// ↳ If z is a root, then z̅ is a root (where z = a+bi & z̅ = a-bi).

[x] Exit text