# Math help

Please help

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#### mmm4444bot

##### Super Moderator
Staff member
Hello. What have you done on paper or thought about so far? Please share.

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!

#### Subhotosh Khan

##### Super Moderator
Staff member
Please help
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

Please share your work/thoughts about this problem.

#### Stouffville

##### New member
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

#### Subhotosh Khan

##### Super Moderator
Staff member
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
Some small (but significant) corrections:
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?

k=11

#### HallsofIvy

##### Elite Member
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
The first line has "= 4" but that has disappeared after.
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.