Question

how can i find the polynomial function with roots 1, 7, and –3 of multiplicity 2 ? my notes dont explain this very well

Answer 1

We are given that the polynomial has roots 1, 7 and -3. Each with multiplicity of 2 so the polynomial can be written as
\[(x-1)^2(x-7)^2(x+3)^2=0\]
Here x-1 is a factor , root 1 Has a multiplicity of 2 so
(x-1)^2 , similarly for other roots.
:)

Answer 2

Did you get wow?

Answer 3

I give you the medal under the assumption that multiplicity 2 is correct.

Answer 4

It's given in the question.

Answer 5

\[(x+1)(x+49)(x+9)\]

Answer 6

would i end up with that?

Answer 7

Use this formula
\[(x-a)^2=x^2-2ax+a^2)\]

Answer 8

\[(x-1)^2=x^2-2x+1\]
do this for all the other terms

Answer 9

i got a 3

Answer 10

How did you get 3?

Answer 11

\[(x^{2}-2(-1)(1) +-1 ^{2})\]

Answer 12

x^2+2+1=3

Answer 13

wow it'll be
\[\huge{ x^2-2(1)(x)+1=x^2-2x+1}\]

Answer 14

\[\huge{(x-2)^2=x^2-2(x)(2)+2^2}\]

Answer 15

Do this for the question, post your answer. I'll help if you are stuck somewhere

Answer 16

is it 4?

Answer 17

nope it's
\[x^2-2(x)(2)+2^2=x^2-4x+4\]

Answer 18

so then what do i do next?

Answer 19

we have
\[(x-1)^2(x-7)^2(x+3)^2=0\]
expand each bracketed term, and then leave, no need for multiplying them, It'll be very long
\[(x^2-2x-1)(x-7)^2(x+3)^2=0\]
expand (x-7)^2 and (x+3)^2 also and that'll be your answer

Answer 20

(x^2 - 14x + 49)(x^2 + 6x + 9) ?

Answer 21

yeah correct:)

Answer 22

thank you

Answer 23

welcome :)