Question

Find the polynomial function with roots 1, 7, and –3 of multiplicity 2

Answer 1

(x - 1)(x - 7)(x + 3)(x + 3) = 0
Can you multiply this out?

Answer 2

you mean foil it?

Answer 3

Yes.

Answer 4

how would i foil 4 terms?

Answer 5

Do by pair: (x + 3)(x + 3)

Answer 6

ohh okay. x^2+3x+3x+9

Answer 7

and the other x^2+7x+7x+7

Answer 8

i messed up sorry .

Answer 9

x^2-7x-x+7

Answer 10

x^2+3x+3x+9 = x² + 6x + 9
Then multiply ( x + 7 ) ( x² + 6x + 9 )

Answer 11

–3 of multiplicity 2 <=> ( x +3)²

Answer 12

Alright. After combining like terms, you get \((x^{2} + 6x + 9)(x^{2} - 8x + 7)\).
Now, in order to multiply trinomials, you use the same concept of FOILing, but maybe this'll help. What I have ^ turns to:
\[x^{2}(x^{2}) + x^{2}(-8x) + x^{2}(7) + 6x(x^{2}) + 6x(-8x) + 6x(7) + 9(x^{2}) + 9(-8x) + 9(7)\]Can you solve this and combine like terms?

Answer 13

\[\small x^{2}(x^{2}) + x^{2}(-8x) + x^{2}(7) + 6x(x^{2}) + 6x(-8x) + 6x(7) + 9(x^{2}) + 9(-8x) + 9(7)\]

Answer 14

x^4-8x^2+6x^2-48x^2+42x+9x^2-72x+63

Answer 15

How'd you get that? Remember the exponent rules. \((x^{a})(x^{b}) = x^{a + b}\)

Answer 16

@Calcmathlete Let me show the easier way

Answer 17

Alright. You're gonna do it a binomial at a time like you were about to before?

Answer 18

yeah, I think the asker stuck with 3 terms

Answer 19

Yeah. it's just expanding the concept of FOILing, so I leave the rest to you :)

Answer 20

Can you multiply
( x + 7 ) ( x² + 6x + 9 ) = ...

Answer 21

( x³ + 6x² + 9x ) + ( 7x² + 42x + 63)
Can you add them together?

Answer 22

Then take the result multiply for ( x -1)

Answer 23

@Jarelii Can you try?

Answer 24

sorry but this just really hard i give up i dont get it