How do you find all rational roots for $4 y^{5} + 8 y^{4} - 29 y^{3} - 42 y^{2} + 45 y + 54 = 0$ $4y^5 + 8y^4 - 29y^3 - 42y^2 + 45y + 54 = 0$ ?

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How do you find all rational roots for $4 y^{5} + 8 y^{4} - 29 y^{3} - 42 y^{2} + 45 y + 54 = 0$ $4y^5 + 8y^4 - 29y^3 - 42y^2 + 45y + 54 = 0$ ?

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Answer 1 · 2016-08-04T20:43:20

Zeros: $- 1, 2, - 3, \frac{3}{2}, - \frac{3}{2}$ $-1, 2, -3, 3/2, -3/2$

Explanation:

$f (y) = 4 y^{5} + 8 y^{4} - 29 y^{3} - 42 y^{2} + 45 y + 54$ $f(y) = 4y^5+8y^4-29y^3-42y^2+45y+54$

By the rational root theorem, any rational zeros of $f (y)$ $f(y)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $54$ $54$ and $q$ $q$ a divisor of the coefficient $4$ $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4}, \pm \frac{1}{2}, \pm \frac{3}{4}, \pm 1, \pm \frac{3}{2}, \pm 2, \pm \frac{9}{4}, \pm 3, \pm \frac{9}{2}, \pm 6, \pm \frac{27}{4}, \pm 9, \pm \frac{27}{2}, \pm 18, \pm 27, \pm 54$ $+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-9/4, +-3, +-9/2, +-6, +-27/4, +-9, +-27/2, +-18, +-27, +-54$

That's rather a lot of possibilities to try, but first note that:

$f (- 1) = - 4 + 8 + 29 - 42 - 45 + 54 = 0$ $f(-1) = -4+8+29-42-45+54 = 0$

So $y = - 1$ $y=-1$ is a zero and $(y + 1)$ $(y+1)$ a factor:

$4 y^{5} + 8 y^{4} - 29 y^{3} - 42 y^{2} + 45 y + 54$ $4y^5+8y^4-29y^3-42y^2+45y+54$

$= (y + 1) (4 y^{4} + 4 y^{3} - 33 y^{2} - 9 y + 54)$ $=(y+1)(4y^4+4y^3-33y^2-9y+54)$

Sticking with simpler calculations, let's try substituting $y = 2$ $y=2$ in the remaining quartic:

$4 y^{4} + 4 y^{3} - 33 y^{2} - 9 y + 54$ $4y^4+4y^3-33y^2-9y+54$

$= 4 (16) + 4 (8) - 33 (4) - 9 (2) + 54$ $=4(16)+4(8)-33(4)-9(2)+54$

$= 64 + 32 - 132 - 18 + 54 = 0$ $=64+32-132-18+54 = 0$

So $y = 2$ $y=2$ is a zero and $(y - 2)$ $(y-2)$ a factor:

$4 y^{4} + 4 y^{3} - 33 y^{2} - 9 y + 54 = (y - 2) (4 y^{3} + 12 y^{2} - 9 y - 27)$ $4y^4+4y^3-33y^2-9y+54 = (y-2)(4y^3+12y^2-9y-27)$

In the remaining cubic, the ratio of the first and second terms is the same as the ratio between the third and fourth terms. Hence it will factor by grouping:

$4 y^{3} + 12 y^{2} - 9 y - 27$ $4y^3+12y^2-9y-27$

$= (4 y^{3} + 12 y^{2}) - (9 y + 27)$ $=(4y^3+12y^2)-(9y+27)$

$= 4 y^{2} (y + 3) - 9 (y + 3)$ $=4y^2(y+3)-9(y+3)$

$= (4 y^{2} - 9) (y + 3)$ $=(4y^2-9)(y+3)$

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

$= (2 y - 3) (2 y + 3) (y + 3)$ $=(2y-3)(2y+3)(y+3)$

Hence three more zeros: $\frac{3}{2}$ $3/2$ , $- \frac{3}{2}$ $-3/2$ , $- 3$ $-3$

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How do you find all rational roots for $4 y^{5} + 8 y^{4} - 29 y^{3} - 42 y^{2} + 45 y + 54 = 0$ $4y^5 + 8y^4 - 29y^3 - 42y^2 + 45y + 54 = 0$ ?

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How do you find all rational roots for 4y5+8y4−29y3−42y2+45y+54=04y^5 + 8y^4 - 29y^3 - 42y^2 + 45y + 54 = 0?

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How do you find all rational roots for $4 y^{5} + 8 y^{4} - 29 y^{3} - 42 y^{2} + 45 y + 54 = 0$ $4y^5 + 8y^4 - 29y^3 - 42y^2 + 45y + 54 = 0$ ?