Find the other zeros of 2x⁴ -3x² +6x²- 2 if - √2 and √2 are zeros of p(x).
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Answer
Given, – √2 and √2 are zeros of p(x).
⇒ (x + √2) and (x – √2) are the factors of p(x).
Thus (x + √2) (x – √2) = (x2 – 2) is also a factor of p(x).
To find the remaining zeros, we find the remaining factors using the division process.
Here, dividend polynomial = p(x) = 2x4 + 7x3 — 8x2 — 14x + 8
and divisor polynomial = s(x) = x2 – 2
Given, – √2 and √2 are zeros of p(x).
⇒ (x + √2) and (x – √2) are the factors of p(x).
Thus (x + √2) (x – √2) = (x2 – 2) is also a factor of p(x).
To find the remaining zeros, we find the remaining factors using the division process.
Here, dividend polynomial = p(x) = 2x4 + 7x3 — 8x2 — 14x + 8
and divisor polynomial = s(x) = x2 – 2
⇒ p(x) = 2x4 + 7x3 — 8x2 — 14x + 8 = (x2 – 2)(2x2 + 7x — 4)
On factorising 2x2 + 7x — 4, we get
2x2 + 7x — 4 = 2x2 + 8x – x — 4
= 2x (x + 4) –1 (x + 4)
= (2x –1) (x + 4)
Hence the other two zeros of p(x) are
and –4.
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