How do you find all rational roots for $8 y^{4} - 6 y^{3} + 17 y^{2} - 12 y + 2 = 0$ $8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0$ ?

Question

How do you find all rational roots for $8 y^{4} - 6 y^{3} + 17 y^{2} - 12 y + 2 = 0$ $8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0$ ?

1 Answer

Impact of this question

1635 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-04T23:02:15

The rational roots are $\frac{1}{4}$ $1/4$ , $\frac{1}{2}$ $1/2$ .

The remaining two roots are $\pm \sqrt{2} i$ $+-sqrt(2)i$

Explanation:

$f (y) = 8 y^{4} - 6 y^{3} + 17 y^{2} - 12 y + 2$ $f(y)=8y^4-6y^3+17y^2-12y+2$

By the rational root theorem, any rational zeros of $f (y)$ $f(y)$ must be expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $2$ $2$ and $q$ $q$ a divisor of the coefficient $8$ $8$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{8}, \pm \frac{1}{4}, \pm \frac{1}{2}, \pm 1, \pm 2$ $+-1/8, +-1/4, +-1/2, +-1, +-2$

In addition, note that $f (- y) = 8 y^{4} + 6 y^{3} + 17 y^{2} + 12 y + 2$ $f(-y) = 8y^4+6y^3+17y^2+12y+2$ has all positive coefficients. Hence $f (y)$ $f(y)$ has no negative zeros.

So the only possible rational zeros of $f (y)$ $f(y)$ are:

$\frac{1}{8}, \frac{1}{4}, \frac{1}{2}, 1, 2$ $1/8, 1/4, 1/2, 1, 2$

We find:

$f (\frac{1}{4}) = 8 {(\frac{1}{4})}^{4} - 6 {(\frac{1}{4})}^{3} + 17 {(\frac{1}{4})}^{2} - 12 (\frac{1}{4}) + 2$ $f(1/4) = 8(1/4)^4-6(1/4)^3+17(1/4)^2-12(1/4)+2$

$= \frac{1 - 3 + 34 - 96 + 64}{32} = 0$ $=(1-3+34-96+64)/32 = 0$

$f (\frac{1}{2}) = 8 {(\frac{1}{2})}^{4} - 6 {(\frac{1}{2})}^{3} + 17 {(\frac{1}{2})}^{2} - 12 (\frac{1}{2}) + 2$ $f(1/2) = 8(1/2)^4-6(1/2)^3+17(1/2)^2-12(1/2)+2$

$= \frac{2 - 3 + 17 - 24 + 8}{4} = 0$ $=(2-3+17-24+8)/4 = 0$

So $y = \frac{1}{4}$ $y=1/4$ and $y = \frac{1}{2}$ $y=1/2$ are zeros and $(4 y - 1)$ $(4y-1)$ and $(2 y - 1)$ $(2y-1)$ are factors:

$8 y^{4} - 6 y^{3} + 17 y^{2} - 12 y + 2$ $8y^4-6y^3+17y^2-12y+2$

$= (4 y - 1) (2 y^{3} - y^{2} + 4 y - 2)$ $=(4y-1)(2y^3-y^2+4y-2)$

$= (4 y - 1) (2 y - 1) (y^{2} + 2)$ $=(4y-1)(2y-1)(y^2+2)$

$y^{2} + 2 \geq 2 > 0$ $y^2+2 >= 2 > 0$ for all Real values of $y$ $y$ , so there are no more Real, let alone rational, zeros.

The last two zeros are $\pm \sqrt{2} i$ $+-sqrt(2)i$ since:

$(y - \sqrt{2} i) (y + \sqrt{2} i) = y^{2} - {(\sqrt{2} i)}^{2} = y^{2} + 2$ $(y-sqrt(2)i)(y+sqrt(2)i) = y^2-(sqrt(2)i)^2 = y^2+2$

Science

Math

Humanities

... and beyond

How do you find all rational roots for $8 y^{4} - 6 y^{3} + 17 y^{2} - 12 y + 2 = 0$ $8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find all rational roots for 8y4−6y3+17y2−12y+2=08y^4 - 6y^3 + 17y^2 - 12y + 2 = 0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find all rational roots for $8 y^{4} - 6 y^{3} + 17 y^{2} - 12 y + 2 = 0$ $8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0$ ?