1. Roots of polynomials

2. FM Roots of polynomials: Related Expressions
KUS objectives
BAT find equations of polynomials from linear transformations of a given polynomial
Starter: Solve
𝑥𝑥𝑥given that one root is 𝑧=1−𝑖
𝑥𝑥𝑥
𝑥=𝑥𝑥𝑥=−2±𝑖

3. Notes
Given the roots of a polynomial, it is possible to find the equation of a second polynomial whose roots are a linear transformation of the first
roots
f(x)
new
roots
f(x-a)

4. WB E1 The cubic equation 𝑥 3 +3 𝑥 2 −8𝑥+6=0 has roots ∝, 𝛽, 𝛾
In shorthand
𝛼= − 𝑏 𝑎
And 𝛼𝛽= 𝑐 𝑎
And 𝛼𝛽𝛾 =− 𝑑 𝑎
WB E1 The cubic equation 𝑥 3 +3 𝑥 2 −8𝑥+6=0
has roots ∝, 𝛽, 𝛾
Find the equations of the polynomials with roots:
∝−1 , 𝛽−1 𝑎𝑛𝑑 (𝛾−1) giving your answer in the
form 𝑤 3 +𝑝 𝑥 2 +𝑞𝑤+𝑟=0
Method 1
𝛼+𝛽+𝛾=− 𝑏 𝑎 =−3
𝛼𝛽+𝛼𝛾+𝛽𝛾= 𝑐 𝑎 =−8
𝛼𝛽𝛾=− 𝑑 𝑎 =−6
Sum 𝛼= = 𝛼+𝛽+𝛾 −3=−6
Pair sum 𝛼𝛽=
𝛼𝛽+𝛼𝛾+𝛽𝛾 −2 𝛼+𝛽+𝛾 +3=1
Product 𝛼−1 𝛽−1 𝛾−1 =
= 𝛼𝛽𝛾 − 𝛼𝛽+𝛼𝛾+𝛽𝛾 + 𝛼+𝛽+𝛾 −1=−2
So the cubic is
𝑎 𝑥 3 −𝑎 𝛼+𝛽+𝛾 𝑥 2 +𝑎 𝛼𝛽+𝛼𝛾+𝛽𝛾 𝑥−𝑎𝛼𝛽𝛾
= 𝒘 𝟑 +𝟔 𝒘 𝟐 +𝒘+𝟐=𝟎

5. WB E1 (cont) The cubic equation 𝑥 3 +3 𝑥 2 −8𝑥+6=0 has roots ∝, 𝛽, 𝛾
In shorthand
𝛼= − 𝑏 𝑎
And 𝛼𝛽= 𝑐 𝑎
And 𝛼𝛽𝛾 =− 𝑑 𝑎
WB E1 (cont) The cubic equation 𝑥 3 +3 𝑥 2 −8𝑥+6=0
has roots ∝, 𝛽, 𝛾
Find the equations of the polynomials with roots:
∝−1 , 𝛽−1 𝑎𝑛𝑑 (𝛾−1) giving your answer in the
form 𝑤 3 +𝑝 𝑥 2 +𝑞𝑤+𝑟=0
Let w=x−1 so 𝑥=𝑤+1
Method 2
Substituting 𝑤 𝑤+1 2 −8 𝑤+1 +6=0
Rearranges to …
𝒘 𝟑 +𝟑 𝒘 𝟐 +𝟑𝒘+𝟏+𝟑 𝒘 𝟐 +𝟐𝒘+𝟏 −𝟖𝒘−𝟖+𝟔=𝟎
Rearranges to 𝒘 𝟑 +𝟔 𝒘 𝟐 +𝒘+𝟐=𝟎

6. WB E2a The cubic equation 𝑥 3 −2 𝑥 2 +3𝑥−4=0 has roots ∝, 𝛽, 𝛾
In shorthand
𝛼= − 𝑏 𝑎
And 𝛼𝛽= 𝑐 𝑎
And 𝛼𝛽𝛾 =− 𝑑 𝑎
WB E2a The cubic equation 𝑥 3 −2 𝑥 2 +3𝑥−4=0
has roots ∝, 𝛽, 𝛾
Find the equations of the polynomials with roots:
a) 2∝, 2𝛽 𝑎𝑛𝑑 2𝛾 b) ∝+3 , 𝛽+3 𝑎𝑛𝑑 (𝛾+3)
a) Method 1
𝛼+𝛽+𝛾=− 𝑏 𝑎 =2
𝛼𝛽+𝛼𝛾+𝛽𝛾= 𝑐 𝑎 =3
𝛼𝛽𝛾=− 𝑑 𝑎 =4
Sum 𝛼 + 2𝛽 + 2𝛾 = 2 𝛼+𝛽+𝛾 =4
Pair sum 2𝛼 2𝛽 + 2𝛼 2𝛾 +(2𝛽) 2𝛾 =
4 𝛼𝛽+𝛼𝛾+𝛽𝛾 =12
Product 2𝛼 2𝛽 2𝛾 =
8 𝛼𝛽𝛾 =32
So the cubic is
𝑎 𝑥 3 −𝑎 𝛼+𝛽+𝛾 𝑥 2 +𝑎 𝛼𝛽+𝛼𝛾+𝛽𝛾 𝑥−𝑎𝛼𝛽𝛾
= 𝒘 𝟑 −𝟒 𝒘 𝟐 +𝟏𝟐𝒘−𝟑𝟐=𝟎

7. WB E2b The cubic equation 𝑥 3 −2 𝑥 2 +3𝑥−4=0 has roots ∝, 𝛽, 𝛾
In shorthand
𝛼= − 𝑏 𝑎
And 𝛼𝛽= 𝑐 𝑎
And 𝛼𝛽𝛾 =− 𝑑 𝑎
WB E2b The cubic equation 𝑥 3 −2 𝑥 2 +3𝑥−4=0
has roots ∝, 𝛽, 𝛾
Find the equations of the polynomials with roots:
a) 2∝, 2𝛽 𝑎𝑛𝑑 2𝛾 b) ∝+3 , 𝛽+3 𝑎𝑛𝑑 (𝛾+3)
b) Method 1
𝛼+𝛽+𝛾=− 𝑏 𝑎 =2
𝛼𝛽+𝛼𝛾+𝛽𝛾= 𝑐 𝑎 =3
𝛼𝛽𝛾=− 𝑑 𝑎 =4
Sum 𝛼+3 + 𝛽+3 + 𝛾+3 = 𝛼+𝛽+𝛾 +9=11
Pair sum 𝛼+3 𝛽+3 + 𝛼+3 𝛾+3 +(𝛽+3) 𝛾+3
= 𝛼𝛽+𝛼𝛾+𝛽𝛾 +6 𝛼+𝛽+𝛾 +27=42
Product 𝛼+3 𝛽+3 𝛾+3 =
= 𝛼𝛽𝛾 +3 𝛼𝛽+𝛼𝛾+𝛽𝛾 +9 𝛼+𝛽+𝛾 ++27=58
So the cubic is
𝑎 𝑥 3 −𝑎 𝛼+𝛽+𝛾 𝑥 2 +𝑎 𝛼𝛽+𝛼𝛾+𝛽𝛾 𝑥−𝑎𝛼𝛽𝛾
= 𝒘 𝟑 −𝟏𝟏 𝒘 𝟐 +𝟒𝟐𝒘−𝟓𝟖=𝟎

8. WB E2 (cont) The cubic equation 𝑥 3 −2 𝑥 2 +3𝑥−4=0 has roots ∝, 𝛽, 𝛾
In shorthand
𝛼= − 𝑏 𝑎
And 𝛼𝛽= 𝑐 𝑎
And 𝛼𝛽𝛾 =− 𝑑 𝑎
WB E2 (cont) The cubic equation 𝑥 3 −2 𝑥 2 +3𝑥−4=0
has roots ∝, 𝛽, 𝛾
Find the equations of the polynomials with roots:
a) 2∝, 2𝛽 𝑎𝑛𝑑 2𝛾 b) ∝+3 , 𝛽+3 𝑎𝑛𝑑 (𝛾+3)
Let w=2x so 𝑥= 𝑤 2
a) Method 2
Substituting 𝑤 −2 𝑤 𝑤 2 −4=0
Multiply through by 8
Rearranges to 𝒘 𝟑 −𝟒 𝒘 𝟐 +𝟏𝟐𝒘−𝟑𝟐=𝟎
b) Method 2
Let w=𝑥+3 so 𝑥=𝑤−3
Substituting 𝑤−3 3 −2 𝑤− 𝑤−3 −4=0
Rearranges to 𝒘 𝟑 −𝟏𝟏 𝒘 𝟐 +𝟒𝟐𝒘−𝟓𝟖=𝟎

9. 𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿 = 𝑐 𝑎 =0
WB E The quartic equation 𝑥 4 −3 𝑥 3 +15𝑥+1=0 has roots ∝, 𝛽, 𝛾, 𝛿 Find the equation with roots: 2∝+1 , 2𝛽+1 , (2𝛾+1)𝑎𝑛𝑑 (2𝛿+1)
𝛼+𝛽+𝛾+𝛿 =− 𝑏 𝑎 =3
In shorthand
𝛼= − 𝑏 𝑎
And 𝛼𝛽= 𝑐 𝑎
And 𝛼𝛽𝛾 =− 𝑑 𝑎
𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿 = 𝑐 𝑎 =0
𝛼𝛽𝛾+𝛼𝛽𝛿+𝛼𝛾𝛿+𝛽𝛾𝛿 =− 𝑑 𝑎 =−15
𝛼𝛽𝛾𝛿 = 𝑒 𝑎 =1
𝛼= 2𝛼+1)+(2𝛽+1)+(2𝛾+1)+(2𝛿+1 =2 𝛼+𝛽+𝛾+𝛿 +4=𝟏𝟎
𝛼𝛽= 2𝛼+1 2𝛽+1 + 2𝛼+1) (2𝛾+1 + 2𝛼+1) (2𝛿+1
2𝛽+1 2𝛾+1 + 2𝛽+1 2𝛿+1 + 2𝛾+1 (2𝛿+1)= …
4 𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿 +6 𝛼+𝛽+𝛾+𝛿 +6
=0+18+6=𝟐𝟒

10. 𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿 = 𝑐 𝑎 =0
WB E3(cont) The quartic equation 𝑥 4 −3 𝑥 3 +15𝑥+1=0 has roots ∝, 𝛽, 𝛾, 𝛿 Find the equation with roots: 2∝+1 , 2𝛽+1 , (2𝛾+1)𝑎𝑛𝑑 (2𝛿+1)
𝛼+𝛽+𝛾+𝛿 =− 𝑏 𝑎 =3
In shorthand
𝛼= − 𝑏 𝑎
And 𝛼𝛽= 𝑐 𝑎
And 𝛼𝛽𝛾 =− 𝑑 𝑎
𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿 = 𝑐 𝑎 =0
𝛼𝛽𝛾+𝛼𝛽𝛿+𝛼𝛾𝛿+𝛽𝛾𝛿 =− 𝑑 𝑎 =−15
𝛼𝛽𝛾𝛿 = 𝑒 𝑎 =1
′ 𝛼𝛽𝛾 ′ =
8 𝛼𝛽𝛾+𝛼𝛽𝛿+𝛼𝛾𝛿+𝛽𝛾𝛿 +8 𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿
+6 𝛼+𝛽+𝛾+𝛿 +4=−𝟗𝟖
2∝+1 2𝛽+1 2𝛾+1 2𝛿+1
=16 𝛼𝛽𝛾𝛿 +8 𝛼𝛽𝛾+𝛼𝛽𝛿+𝛼𝛾𝛿+𝛽𝛾𝛿 +4 𝛼𝛽+𝛼𝛾+𝛼𝛿+𝛽𝛾+𝛽𝛿+𝛾𝛿
+2 𝛼+𝛽+𝛾+𝛿 +1=−𝟗𝟕
So the quartic is
=𝑎 𝑥 4 −𝑎 𝛼+𝛽+𝛾+𝛿 𝑥 3 +𝑎 𝜶𝜷+𝜶𝜸+𝜶𝜹+𝜷𝜸+𝜷𝜹+𝜸𝜹 𝑥 2
𝑎 𝛼𝛽𝛾+𝛼𝛽𝛿+𝛼𝛾𝛿+𝛽𝛾𝛿 𝑥+𝑎𝛼𝛽𝛾𝛿
= 𝒘 𝟒 −𝟏𝟎 𝒘 𝟑 +𝟐𝟒 𝒘 𝟐 +𝟗𝟖𝒘−𝟗𝟕=𝟎

11. Substituting 𝑤−1 2 4 −3 𝑤−1 2 3 +15 𝑤−1 2 +1=0
WB E3 (cont) The quartic equation 𝑥 4 −3 𝑥 3 +15𝑥+1=0 has roots ∝, 𝛽, 𝛾, 𝛿 Find the equation with roots: 2∝+1 , 2𝛽+1 , (2𝛾+1)𝑎𝑛𝑑 (2𝛿+1)
Let w=2x+1 so 𝑥= 𝑤−1 2
Method 2
Substituting 𝑤− −3 𝑤− 𝑤− =0
Multiply through by 16 Rearranges to …
𝒘 𝟒 −𝟒𝒘 𝟑 +𝟔 𝒘 𝟐 −𝟒𝒘+𝟏−𝟔 𝒘 𝟑 −𝟑𝒘 𝟐 +𝟑𝒘−𝟏 +𝟏𝟐𝟎𝒘−𝟏𝟐𝟎+𝟏𝟔=𝟎
Rearranges to 𝒘 𝟒 −𝟏𝟎𝒘 𝟑 +𝟐𝟒 𝒘 𝟐 +𝟗𝟖𝒘−𝟗𝟕=𝟎

12. NOW DO Ex 4D self-assess One thing learned is –
KUS objectives
BAT find equations of polynomials from linear transformations of a given polynomial
NOW DO Ex 4D
self-assess
One thing learned is –
One thing to improve is –

13. END

14. 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 𝑦− 𝑑𝑦
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑦−1 𝑑𝑦
= 𝜋 𝑦 2 −𝑦 8 4
= 𝜋 64−8 −𝜋 16−4 =44𝜋