Show that the given value(s) of c are zeroes of P(x), and find all other zeroes of P(x).?

Question

P(x)=3x^4-x^3-21x^2-11x+6, c=1/3,-2

Answer 1

P(1/3)=3(1/3)^4-(1/3)^3-21(1/3)^2-11(1/3)+6
P(1/3)=3(1/81)-(1/27)-21(1/9)-11(1/3)+6
P(1/3)=(1/27)-(1/27)-(7/3)-(11/3)+6
P(1/3)=0

P(-2)=3(-2)^4-(-2)^3-21(-2)^2-11(-2)+6
P(-2)=3(16)-(-8)-21(4)-11(-2)+6
P(-2)=48+8-84+22+6
P(-2)=0

Now factor (x+2) and (x-1/3) out of the original equation (I used synthetic division). That leaves you with:
3x^2-6x-9=0
3(x^2-6x-9)=0
3(x-3)(x+1)=0

So, the other two zeros are -1 and 3.

Answer 2

Plug x = 1/3 into P(x) and you will get P(1/3)=0. Plug x = -2 into P(x) and you will get P(x)=0. THis means that 1/3 and -2 are indeed zeroes of P(x). It also means that (x-1/3) and (x+2) are factors of P(x).So divide P(x) by (x-1/3)(x+2) and you will get a quadratic function. You can then use the quadratic formula to find the other two roots (zeroes) of P(x).

Answer 3

p(x)=3x^4-x^3-21x^2-11x+6
p(1/3)= 3(1/81)-(1/27) -21(1/9) -11(1/3) +6
=1/27 -1/27 -7/3 -11/3 +6 =0
similarly p(-2)=0

by synthetic division
1/3 | 3 -1 -21 -11 6
........0 1.. 0 ..-7. -6
________________
.-2/..3 0. -21 -18 0
...... 0 -6 12 +18
_______________
.......3 -6 -9 0
I.E,
3x^2-6x-9=0
x^2-2x-3=0
(x-3)(x-1)=0
x=3,1,1/3,-2

Answer 4

use the values of c as values instedad of x, do it for both of them and then... find the product and the sum of the rrots and solve for the other 2, i wont do it but maybe theres a double root.