Asked by Vincent
Find all the zeros of the function and write the polynomial as a product of linear factors.
g(x)=3x^3-4x^2+8x+8
On my graphing calculator it says that it is -2/3. But when I do it by hand using synthetic division, I don't get a zero. I'm going crazy here... I've tried -2/3 and 2/3 none of them work... But when I look on my graphing calculator it says that it is -2/3. Help?
g(x)=3x^3-4x^2+8x+8
On my graphing calculator it says that it is -2/3. But when I do it by hand using synthetic division, I don't get a zero. I'm going crazy here... I've tried -2/3 and 2/3 none of them work... But when I look on my graphing calculator it says that it is -2/3. Help?
Answers
Answered by
Reiny
if x= -2/3 then one of the factors is 3x+2
I did algebraic long division and got x^2 - 2x + 4 with no remainder.
so g(x)=3x^3-4x^2+8x+8
= (3x+2)(x^2-2x+4)
so x= -2/3 or x = 1 ± √-3 wich are complex numbers
you will not be able to express it as a product of only linear factors, but rather as a
product of a linear factor times a quadratic factor
BTW, I did get a zero when doing synthetic division.
(-2/3) │ 3 -4 8 8
********** -2 4 -8
******** 3 -6 12 0
I did algebraic long division and got x^2 - 2x + 4 with no remainder.
so g(x)=3x^3-4x^2+8x+8
= (3x+2)(x^2-2x+4)
so x= -2/3 or x = 1 ± √-3 wich are complex numbers
you will not be able to express it as a product of only linear factors, but rather as a
product of a linear factor times a quadratic factor
BTW, I did get a zero when doing synthetic division.
(-2/3) │ 3 -4 8 8
********** -2 4 -8
******** 3 -6 12 0