What are all the possible rational zeros for $f (x) = 3 x^{3} + 11 x^{2} + 5 x - 3$ $f(x)=3x^3+11x^2+5x-3$ and how do you find all zeros?

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Answer 1 · 2016-09-07T14:21:54

The Zeroes of $f$ $f$ are $\frac{1}{3}, - 3, &, - 1$ $1/3,-3,&,-1$ .

We will factorise $f (x) = 3 x^{3} + 11 x^{2} + 5 x - 3,$ $f(x)=3x^3+11x^2+5x-3,$ to find its zeroes.

Observe that,

The Sum of the co-effs. of Odd-powered terms $= 3 + 5 = 8,$ $=3+5=8,$ and,

that of the Even-powered ones $= 11 - 3 = 8$ $=11-3=8$ .

Hence, $(x + 1)$ $(x+1)$ is a factor of $f (x)$ $f(x)$ .

Now, $f (x) = 3 x^{3} + 11 x^{2} + 5 x - 3$ $f(x)=3x^3+11x^2+5x-3$

$=ul(3x^3+3x^2)+ul(8x^2+8x)-ul(3x-3)$

$=3x^2(x+1)+8x(x+1)-3(x+1)$

$=(x+1)(3x^2+8x-3)$

$=(x+1){ul(3x^2+9x)-ul(x-3)}$

$=(x+1){3x(x+3)-1(x+3)}$

$=(x+1)(x+3)(3x-1)$

Hence, the Zeroes of $f$ are $1/3,-3,&,-1$ .

What are all the possible rational zeros for f(x)=3x^3+11x^2+5x-3f(x)=3x3+11x2+5x−3f(x)=3x^3+11x^2+5x-3 and how do you find all zeros?