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

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Answer 1 · 2016-09-11T10:43:44

the zeroes are. $- 1, - 1, - \frac{1}{2}$ $-1,-1,-1/2$ .

$f (x) = 2 x^{3} + 5 x^{2} + 4 x + 1$ $f(x)=2x^3+5x^2+4x+1$

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

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

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

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. $-1,-1,-1/2$ .

What are all the possible rational zeros for f(x)=2x^3+5x^2+4x+1f(x)=2x3+5x2+4x+1f(x)=2x^3+5x^2+4x+1 and how do you find all zeros?