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Question
Find all the zeros of the polynomial x3 + 3x2 – 2x – 6, if two of its zeros are `-sqrt(2)` and `sqrt(2)`.
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Solution
We know that if x = a is a zero of a polynomial, then x – a is a factor of f(x).
Since, `sqrt2` and `-sqrt2` are zeros of f(x).
Therefore
`(x+sqrt2)(x-sqrt2)=x^2-(sqrt2)^2`
= x2 – 2
x2 – 2 is a factor of f(x). Now, we divide x3 + 3x2 – 2x – 6 by g(x) = x2 – 2 to find the zero of f(x).
x + 3
`x^2 - 2")"overline(+ \cancel(x^3) + 3x^2 - \cancel(2x) - 6)`
`+ \cancel(x^3) - 0 - \cancel(2x)`
– +
`+ \cancel(3x^2) - \cancel(6)`
`+ \cancel(3x^2) - \cancel(6)`
– +
0
By using division algorithm we have
f(x) = g(x) × q(x) – r(x)
x3 + 3x2 – 2x – 6 = (x2 – 2)(x + 3) – 0
x3 + 3x2 − 2x − 6 `=(x+sqrt2)(x-sqrt2)(x+3)`
Hence, the zeros of the given polynomials are `-sqrt2, +sqrt2` and –3.
