#### Stouffville

##### New member

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- Thread starter Stouffville
- Start date

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If you're not sure what to try, then substitute x=2 (the given root), simplify the resulting polynomial, and post what you get. Thank you!

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here what I have done

3x^4 - (k-2)x^3 + (k+3)2=4

48-8k -16+2k + 6 -4

-8k + 2k + 48-16+6-4

-6k + 34

3x^4 - (k-2)x^3 + (k+3)2=4

48-8k -16+2k + 6 -4

-8k + 2k + 48-16+6-4

-6k + 34

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Some small (but significant) corrections:here what I have done

3x^4 - (k-2)x^3 + (k+3)2=4

48-8k -16+2k + 6 -4

-8k + 2k + 48-16+6-4

-6k + 34

3x^4 - (k-2)x^3 + (k+3)2- 4

48-8k+16+2k + 6 -4

-8k + 2k + 48+16+6-4

-6k + 66

What should be the value of the function - while evaluated at one of the roots?

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k=11

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The first line has "= 4" but that has disappeared after.here what I have done

3x^4 - (k-2)x^3 + (k+3)2=4

48-8k -16+2k + 6 -4

-8k + 2k + 48-16+6-4

-6k + 34

Also you still have "x" everywhere except at the last "(k+ 3)x"

Your original equation was 3x^4- (k-2)x^3+ (k+3)x- 4= 0.

Since x= 2 is a root, setting x= 2 we have

3(16)- (k- 2)(8)+ (k+ 3)(2)- 4= 0

48- 8k+ 16+ 2k+ 6- 4= 0

(48+ 16+ 6-4)- (8- 2)k= 0

66- 6k= 0

6k= 66

k= 66/6= 11.