Question

Find all the zeroes of the polynomial x^4-5x^3+2x^2+10x-8, if two of its zeroes are √2 and -√2​

Answer 1

Answer:

All the zeroes of the polynomial x⁴ – 5x³ + 2x² + 10x – 8 are: √2, –√2, 1, 4.

Step-by-step explanation:

Given,

p(x) = x⁴ – 5x³ + 2x² + 10x – 8

As √2 and –√2 are zeroes,

(x – √2)(x + √2) = x²– 2 is a factor.

$\frac{ {x}^{4} - 5 {x}^{3} + 2 {x}^{2} + 10x - 8}{(x² - 2)} = {x}^{2} - 5x + 4$

(See attachment for steps of division.)

Now, x² – 5x + 4

= x² – x – 4x + 4 (Splitting the middle term)

= x (x–1) – 4 (x–1)

= (x–1)(x–4)

So, x⁴ – 5x³ + 2x² + 10x – 8

= (x – √2)(x + √2) (x² – 5x + 4)

= (x – √2) (x + √2) (x – 1) (x – 4)

Hence, all the zeroes of p(x) are:

x – √2 = 0 ⇒ x = √2

x + √2 = 0 ⇒ x = –√2

x – 1 = 0 ⇒ x = 1

x – 4 = 0 ⇒ x = 4

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