# Math help

#### 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! #### 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?

#### Stouffville

##### New member
k=11

• Jomo

#### 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.