How do you find the zeros of $f (x) = 7 x^{3} + 4 x^{2} - 3 x + 3$ $f(x) = 7x^3 + 4x^2 - 3x + 3$ ?

Question

How do you find the zeros of $f (x) = 7 x^{3} + 4 x^{2} - 3 x + 3$ $f(x) = 7x^3 + 4x^2 - 3x + 3$ ?

1 Answer

Impact of this question

1633 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-19T21:55:22

Use a substitution to get into the form $t^{3} + p t + q = 0$ $t^3+pt+q = 0$ then Cardano's method...

Explanation:

When a cubic has one irrational Real zero and two non-Real Complex zeros, I like to solve using Cardano's method.

First, I would like to have a cubic with integer coefficients and no square term. To get there from our $f (x)$ $f(x)$ , multiply by $3^{3} \cdot 7^{2} = 1323$ $3^3*7^2 = 1323$ and substitute $t = 21 x + 4$ $t = 21x+4$ as follows:

$1323 (7 x^{3} + 4 x^{2} - 3 x + 3)$ $1323(7x^3+4x^2-3x+3)$

$= 9261 x^{3} + 5292 x^{2} - 3969 x + 3969$ $=9261x^3+5292x^2-3969x+3969$

$= (9261 x^{3} + 5292 x^{2} + 1008 x + 64) - (4977 x + 948) + 4853$ $=(9261x^3+5292x^2+1008x+64)-(4977x+948)+4853$

$= (21^{3} x^{3} + (3 \cdot 21^{2} \cdot 4) x^{2} + (3 \cdot 21 \cdot 4^{2}) x + 4^{3}) - ((237 \cdot 21) x + (237 \cdot 4)) + 4853$ $=(21^3x^3+(3*21^2*4)x^2+(3*21*4^2)x+4^3)-((237*21)x+(237*4))+4853$

$= {(21 x + 4)}^{3} - 237 (21 x + 4) + 4853$ $=(21x+4)^3-237(21x+4)+4853$

$= t^{3} - 237 t + 4853$ $=t^3-237t+4853$

So we want to solve $t^{3} - 237 t + 4853 = 0$ $t^3-237t+4853 = 0$

Next substitute $t = u + v$ $t = u + v$ to get:

$0 = {(u + v)}^{3} - 237 (u + v) + 4853$ $0 = (u+v)^3-237(u+v)+4853$

$= u^{3} + v^{3} + (3 u v - 237) (u + v) + 4853$ $=u^3+v^3+(3uv-237)(u+v)+4853$

$= u^{3} + v^{3} + 3 (u v - 79) (u + v) + 4853$ $=u^3+v^3+3(uv-79)(u+v)+4853$

Add the constraint $v = \frac{79}{u}$ $v = 79/u$ and multiply through by $u^{3}$ $u^3$ to get:

${(u^{3})}^{2} + 4853 (u^{3}) + 493039 = 0$ $(u^3)^2+4853(u^3)+493039 = 0$

Use the quadratic formula to find:

$u^{3} = \frac{- 4853 \pm \sqrt{4853^{2} - 4 \cdot 1 \cdot 493039}}{2}$ $u^3 = (-4853+-sqrt(4853^2-4*1*493039))/2$

$= \frac{- 4853 \pm \sqrt{21579453}}{2}$ $=(-4853+-sqrt(21579453))/2$

$= \frac{- 4853 \pm 63 \sqrt{5437}}{2}$ $=(-4853+-63 sqrt(5437))/2$

Since the derivation was symmetric in $u$ $u$ and $v$ $v$ , one of these roots can be taken as $u^{3}$ $u^3$ and the other $v^{3}$ $v^3$ to give us the Real root:

$t_{1} = \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}$ $t_1 = root(3)((-4853+63 sqrt(5437))/2)+root(3)((-4853-63 sqrt(5437))/2)$

and the Complex roots:

$t_{2} = ω \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + ω^{2} \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}$ $t_2 = omega root(3)((-4853+63 sqrt(5437))/2)+ omega^2 root(3)((-4853-63 sqrt(5437))/2)$

$t_{3} = ω^{2} \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + ω \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}$ $t_3 = omega^2 root(3)((-4853+63 sqrt(5437))/2)+ omega root(3)((-4853-63 sqrt(5437))/2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ $1$

Then $x = \frac{t - 4}{21}$ $x = (t-4)/21$ , so the zeros of $f (x)$ $f(x)$ are:

$x_{1} = \frac{1}{21} ⎛ ⎝ - 4 + \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}} ⎞ ⎠$ $x_1 = 1/21(-4+ root(3)((-4853+63 sqrt(5437))/2)+root(3)((-4853-63 sqrt(5437))/2))$

$x_{2} = \frac{1}{21} ⎛ ⎝ - 4 + ω \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + ω^{2} \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}} ⎞ ⎠$ $x_2 = 1/21(-4+ omega root(3)((-4853+63 sqrt(5437))/2)+ omega^2 root(3)((-4853-63 sqrt(5437))/2))$

$x_{3} = \frac{1}{21} ⎛ ⎝ - 4 + ω^{2} \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + ω \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}} ⎞ ⎠$ $x_3 = 1/21(-4+ omega^2 root(3)((-4853+63 sqrt(5437))/2)+ omega root(3)((-4853-63 sqrt(5437))/2))$

Science

Math

Humanities

... and beyond

How do you find the zeros of $f (x) = 7 x^{3} + 4 x^{2} - 3 x + 3$ $f(x) = 7x^3 + 4x^2 - 3x + 3$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the zeros of f(x)=7x3+4x2−3x+3f(x) = 7x^3 + 4x^2 - 3x + 3?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the zeros of $f (x) = 7 x^{3} + 4 x^{2} - 3 x + 3$ $f(x) = 7x^3 + 4x^2 - 3x + 3$ ?