Question

Find all zeros of the polynomial 2x⁴ + 7x³ – 19x² – 14x + 30, if two of its zeros are `sqrt(2)` and `-sqrt(2)`.

Answer 1

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 2x⁴ + 7x³ – 19x² – 14x + 30 by g(x) = x² – 2 to find the zero of f(x).

                    2x² + 7x – 15
`x^2 - 2")"overline(+ \cancel(2x^4) + 7x^3 - 19x^2 - 14x + 30)`
            `+ \cancel(2x^4) + 0 - 4x^2`
             –         –      +        
                       `+ \cancel(7x^3) - 15x^2 - \cancel(14x)`
                       `+ \cancel(7x^3) - 0       - \cancel(14x)`
                       –           +          +          
                                 `- \cancel(15x^2) + \cancel(30)`
                                 `- \cancel(15x^2) + \cancel(30)`
                                  +            –          
                                            0

By using division algorithm we have

f(x) = g(x) × q(x) – r(x)

2x⁴ + 7x³ – 19x² – 14x + 30 = (x² – 2)(2x² + 7x – 15) + 0

2x⁴ + 7x³ – 19x² – 14x + 30 `=(x+sqrt2)(x-sqrt2)(2x^2+10x-3x-15)`

2x⁴ + 7x³ – 19x² – 14x + 30 `=(x+sqrt2)(x-sqrt2)[2x(x+5)-3(x+5)]`

2x⁴ + 7x³ – 19x² – 14x + 30 `=(x+sqrt2)(x-sqrt2)(2x-3)(x+5)`

Hence, the zeros of the given polynomial are `-sqrt2, +sqrt2, (+3)/2, -5`.