I think I know how to go about this question - using synthetic division / nested form of a polynomial.
The problem is, I cannot seem to get the correct answer.
I believe the graph cuts the x-axis (roots) at (3,0) and (-3,0), but cannot get this answer through factorizing the equation.
Please help!!
Factorise Fully; x^4-6x^2-27?
2008-10-19 9:34 am
回答 (5)
2008-10-19 9:45 am
✔ 最佳答案
The expression you have is called a quadratic in x², so if you let p = x² you can rewrite is as:p² - 6p - 27
This can be factorized as
(p - 9) (p + 3)
So replacing p = x²
(x² - 9) (x² + 3)
or
(x - 3)(x + 3) (x² + 3)
If you need to find the x-intercepts, equate this to zero and you will get:
x - 3 = 0; x = 3
x + 3 = 0; x = -3
So the points are (3,0) and (-3,0)
Equating (x² + 3) to zero will only give the complex roots which you cannot plot.
2008-10-19 4:43 pm
(x^2 -9)(x^2 +3)=(x-3)(x+3)(x^2 +3)
2008-10-19 4:42 pm
x⁴ - 6x² - 27 = (x² - 9)(x² + 3)
. . . . . . . . . = (x - 3)(x + 3)(x² + 3)
. . . . . . . . . = (x - 3)(x + 3)(x² + 3)
2008-10-19 4:41 pm
x^4 - 6x^2 - 27
(x^2 + 3)(x^2 - 9)
(x^2 + 3)(x + 3)(x - 3)
I hope you can now get the correct answer
(x^2 + 3)(x^2 - 9)
(x^2 + 3)(x + 3)(x - 3)
I hope you can now get the correct answer
2008-10-19 4:40 pm
x^4 - 6x^2 - 27
= x^4 + 3x^2 - 9x^2 - 27
= (x^4 + 3x^2) - (9x^2 + 27)
= x^2(x^2 + 3) - 9(x^2 + 2)
= (x^2 + 3)(x^2 - 9)
= (x^2 + 3)(x + 3)(x - 3)
= x^4 + 3x^2 - 9x^2 - 27
= (x^4 + 3x^2) - (9x^2 + 27)
= x^2(x^2 + 3) - 9(x^2 + 2)
= (x^2 + 3)(x^2 - 9)
= (x^2 + 3)(x + 3)(x - 3)
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