Question

factorise x³-1155x²+354654x-19300680 = 0​

Answer 1

Answer:

Let's start with a simple-ish quadratic expression, such as x 2 − 9 x + 18 . The way to factorise is to find two numbers that multiply together to make 18 but add to make -9. Eighteen doesn't have all that many factor pairs - (1, 18), (2, 9), (3,6) and their negative counterparts.

Answer 2

Answer

$x^3 - 1155 x^2 + 354654 x - 19300680 = ( x - 666 )( x - 420 )( x - 69 )$

Below is a way to approximate the solution.

• $54.9 \mathrm{(approx.)} < x < 1155$

 

-Answer by Takenname. Copying is spamming.-

 

Approximation

(For comparison.)

$1155 = 3 \times 5 \times 7 \times 11$

$354654 = 2 \times 3^2 \times 17 \times 61$

$19300680 = 2^3 \times 3^4 \times 5 \times 7 \times 23 \times 37$

 

Set a new parameter to eliminate factor 3.

Let $x=3t$.

$(3t)^3-1155(3t)^2+354654(3t)-19300680=0$

$\implies (27t^3) - 1155(9t^2) + 354654(3t) - 19300680 = 0$

$\implies t^3 - 385t^2 + 39046t - 714840 = 0$

 

(For comparison.)

$385 = 5 \times 7 \times 11$

$39046 = 2 \times 17 \times 19 \times 61$

$714840 = 2^3 \times 3 \times 5 \times 7 \times 23 \times 37$

 

Parity

$\implies t^3 - 385t^2 + 2 ( 19523t - 357420 ) = 0$

Based on the parity, $t^3-385t^2$ must be even.

$t^3 - 385 t^2 = t^2 ( t - 385 )$, but this is always even, so parity is rejected.

Sign

The signs should be different as well.

• $t < 385$ then $t^2( t - 385 ) < 0$.
• $t > \dfrac{357420}{19523}$ then $2( 19523t - 1057260 ) > 0$.

 

The solutions will lie in $\dfrac{357420}{19523} < t < 385$. The number of real solutions is 1 or 3, according to the graph.

 

For your information. (Attachment included.)

Also, the derivative method can be used here. The derivative is 'slope'. In a linear function, the slope is positive then the value gets greater.

For example, if the derivative is positive, the value gets greater. If the function is positive there, there is no chance of zeros existing.

 

Solving the equation gives $t=22$, $t=140$ or $t=222$.