Roots of polynomials

FM Roots of polynomials: Related Expressions KUS objectives BAT Evaluate expressions related to the roots of polynomials Starter: Solve 𝑥𝑥𝑥given that one root is 𝑧=1−𝑖 𝑥𝑥𝑥 𝑥=𝑥𝑥𝑥=−2±𝑖

Notes Spot the patterns I 1 𝛼 + 1 𝛽 = 𝛼+𝛽 𝛼𝛽 Rules for reciprocals 1 𝛼 + 1 𝛽 + 1 𝛾 = 𝛼𝛽+𝛼𝛾+𝛽𝛾 𝛼𝛽𝛾 1 𝛼 + 1 𝛽 + 1 𝛾 + 1 𝛿 = 𝛼𝛽𝛾+𝛼𝛽𝛿+𝛼𝛾𝛿+𝛽𝛾𝛿 𝛼𝛽𝛾𝛿 Rules for powers 𝛼 𝑛 × 𝛽 𝑛 = 𝛼𝛽 𝑛 𝛼 𝑛 × 𝛽 𝑛 × 𝛾 𝑛 = 𝛼𝛽𝛾 𝑛 𝛼 𝑛 × 𝛽 𝑛 × 𝛾 𝑛 × 𝛿 𝑛 = 𝛼𝛽𝛾𝛿 𝑛

Rules for roots of polynomials used earlier in this topic Notes Spot the patterns II Rules for roots of polynomials used earlier in this topic 𝛼 2 × 𝛽 2 = 𝛼+𝛽 2 −2𝛼𝛽 𝛼 3 + 𝛽 3 = 𝛼+𝛽 3 −3𝛼𝛽 𝛼+𝛽 There are equivalent results for higher powers We can use these to find expressions for sums of squares and sums of cubes

𝑎) 𝛼+𝛽+𝛾 2 = 𝛼 2 +𝛼𝛽+𝛼𝛾+𝛽𝛼+ 𝛽 2 +𝛽𝛾+𝛾𝛼+𝛾𝛽+ 𝛾 2 WB D1 Sums of squares Expand 𝛼+𝛽+𝛾 2 A cubic equation has roots 𝛼, 𝛽, 𝛾 such that 𝛼𝛽+𝛽𝛾+𝛼𝛾=7 and 𝛼+𝛽+𝛾=−3 Find the value of 𝛼 2 + 𝛽 2 + 𝛾 2 𝑎) 𝛼+𝛽+𝛾 2 = 𝛼 2 +𝛼𝛽+𝛼𝛾+𝛽𝛼+ 𝛽 2 +𝛽𝛾+𝛾𝛼+𝛾𝛽+ 𝛾 2 𝛼+𝛽+𝛾 2 = 𝛼 2 + 𝛽 2 + 𝛾 2 +2 𝛼𝛽+𝛽𝛾+𝛼𝛾 𝑏) 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑔𝑖𝑣𝑒𝑠 −3 2 = 𝛼 2 + 𝛽 2 + 𝛾 2 +2(7) 𝛼 2 + 𝛽 2 + 𝛾 2 =−5

Notes Spot the patterns III Rules for sums of squares Quadratic 𝛼 2 + 𝛽 2 = 𝛼+𝛽 2 −2 𝛼𝛽 Cubic 𝛼 2 + 𝛽 2 + 𝛾 2 = 𝛼+𝛽+𝛾 2 −2 𝛼𝛽+𝛽𝛾+𝛼𝛾 Quartic 𝛼 2 + 𝛽 2 + 𝛾 2 + 𝛿 2 = 𝛼+𝛽+𝛾+𝛿 2 −2 𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿 Rules for sums of cubes Quadratic 𝛼 3 + 𝛽 3 = 𝛼+𝛽 3 −3𝛼𝛽 𝛼+𝛽 Cubic 𝛼 3 + 𝛽 3 + 𝛾 3 = 𝛼+𝛽+𝛾 3 −3 𝛼+𝛽+𝛾 𝛼𝛽+𝛽𝛾+𝛼𝛾 +3𝛼𝛽𝛾

NOW DO Ex 4D WB D2 The three roots of a cubic equation are 𝛼, 𝛽 ,𝛾 Given that 𝛼𝛽𝛾=4; 𝛼𝛽+𝛽𝛾+𝛼𝛾=−5 and 𝛼+𝛽+𝛾=3 Find the value of (𝛼+3)(𝛽+3)(𝛾+3) 𝛼+3 𝛽+3 𝛾+3 = =𝛼𝛽𝛾+3𝛼𝛽+3𝛼𝛾+9𝛼+3𝛽𝛾+9𝛽+9𝛾+27 =𝛼𝛽𝛾+3 𝛼𝛽+𝛼𝛾+𝛽𝛾 +9(𝛼+𝛽+𝛾)+27 =4 +3 −5 +9(3)+27 = 43 NOW DO Ex 4D

One thing to improve is – KUS objectives BAT Evaluate expressions related to the roots of polynomials self-assess One thing learned is – One thing to improve is –

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WB 6 The region R is bounded by the curve 𝑥= 2𝑦−1 , the y-axis and the vertical lines y=4 and y = 8 Find the volume of the solid formed when the region is rotated 2π radians about the y-axis. Give your answer as a multiple of π 𝑥= 2𝑦−1 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 4 8 2𝑦−1 2 𝑑𝑦 𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 4 8 2𝑦−1 𝑑𝑦 = 𝜋 𝑦 2 −𝑦 8 4 = 𝜋 64−8 −𝜋 16−4 =44𝜋