Question

What are all the possible rational zeros for `f(x)=2x^3+5x^2+4x+1` and how do you find all zeros?

Answer 1

the zeroes are. `-1,-1,-1/2`.

Explanation:

`f(x)=2x^3+5x^2+4x+1`

Note that, the sum of the co-effs. of odd-powered terms of

`x`=2+4=6, and, that of even-powered`=5+1=6.`.

`:. (x+1)" is a factor of "f(x)`.

Now, `f(x)=2x^3+5x^2+4x+1`

`=ul(2x^2+2x^2)+ul(3x^2+3x)+ul(x+1)`

`=2x^2(x+1)+3x(x+1)+1(x+1)`

`=(x+1)(2x^2+3x+1)`

`(x+1){ul(2x^2+2x)+ul(x+1)}`

`=(x+1){2x(x+1)=1(x+1)}`

`=(x+1){(x+1))(2x+1)}`

`=(x+1)^2(2x+1)`

Hence, the zeroes are. `-1,-1,-1/2`.