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I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\ (x+1)^5 + 36(x+1) = 13 (x +1)^3\\ (x+1)^4 +36 = 13 (x+1)^2 $$

But, don't understand how to solve further. Can somebody show step by step please. Thanks!

  • 0
    let $(x+1)^2=t$ – 2017-02-02
  • 0
    set $y = (x+1)^2$ – 2017-02-02
  • 0
    $36(x+1)$ then factor $(x+1)$ on LHS and RHS and group all in LHS – 2017-02-02
  • 4
    Your last step is invalid if $x=-1$. – 2017-02-02

5 Answers 5

4

Hint:

$$(x+1)^5-13(x+1)^3+36(x+1)=0$$ $$\left[(x+1)^4-13(x+1)^2+36\right](x+1)=0$$ $$\left((x+1)^2-9\right)\left((x+1)^2-4\right)(x+1)=0$$

2

Let $$(x+1)=t$$ Then $t$ satisfies the quintic $$t^5+36t=13t^3 \iff t(t^2-4)(t^2-9)=0 \iff t(t-2)(t+2)(t-3)(t+3)=0$$ Actually, in what you did your last step was invalid as you did not consider the case when you have that $x+1=0$.

2

It's a simple problem. Well $$(x+1)^5+36x+36=13(x+1)^3$$

$$\implies (x+1)^5–13(x+1)^3+36(x+1) = 0$$

Let $y=x+1$

$$\implies y^5-13y^3+36y = 0$$

$$\implies y(y^4-13y^2+36)= 0$$

$$\implies y(y^2-9)(y^2-4)= 0$$ $$\implies y = 0,\pm 2,\pm 3$$ $$\implies x+1 = 0,\pm 2,\pm 3$$ $$\implies x = -4, -3, -1, 1, 2$$

and we’re done.

0

Continue from your last step.

Put $(x + 1)^2 = z$

We have,

$z^2 + 36 = 13z$

$z^2 - 13z + 36 = 0$

$z^2 - 9z - 4z + 36 = 0$

$z(z - 9) - 4(z - 9) = 0$

$(z - 9)(z - 4) = 0$

When z = 9

$(x + 1)^2 = 9$

$(x + 1) = \pm 3$

x = 2 or -4

When z = 4

$(x + 1)^2 = 4$

$(x + 1) = \pm 2$

x = 1 or -3

x = -4, -3, 1, 2

0

Assuming that you don't see the direct substitution $t=x+1$ for whatever reason, note that $(x+1)^5 +36x+36 - 13(x+1)^3$ is a monic polynomial with the free term $1+36-13=24\,$. By the rational root theorem, try the divisors of $24$ for possible rational roots, and enjoy your luck.