3 and - 3 are zeroes of f(x) = x4 – 3x3 – x2 + 9x – 6. Find all the zeroes of p(x).
Answers
Answer:
x= ± 3 [SQUARING BOTH SIDE ]
(x)^2=(3)^2
x^2=9
x^2-9=0
x² - 3x + 2
----------------------
x^2-9|| x4 – 3x3 – x2 + 9x – 6
x^4 -3x²
(-) (+)
--------------------------
0-3x³+2x²+9x-6
-3x³ +9x
( +) ( -)
----------------------------
0+2x²+0-6
2x² -6
( - ) ( +)
---------------------------
0 0
=>x^4-3x³-x²+9x-6
= (x²-3x+2)(x²- 3)
=>(x²-3x+2)
=( x²-2x-x+2)
=x(x-2)-1(x-2)
=(x-2)(x-1)
=>x=2,1
=>x^4-3x³-x²+9x-6
= (x-2)(x-1)(x+√3)(x-√3)
Therefore the zeros are 2,1,√3 and -√3
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