Question

For which values of a and b, are the zeroes of q(x) = x³ + 2x² + a also the zeroes of the polynomial p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x + b? Which zeroes of p(x) are not the zeroes of q(x)?

Answer 1

Since zeroes of q(x) = x² + 2x² + a are also the zeroes of the polynomial

p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x + b

So, q(x) is a factor of p(x).

Then, we use a division algorithm.

                         `x^3 - 3x + 2`
`x^3 + 2x^2 + a")"overline(x^5 - x^4 - 4x^3 + 3x^2 + 3x + b)`
                       x⁵ + 2x⁴             + ax²
                      (–)  (–)                (–)                        
                            –3x⁴ – 4x³ + (3 – a)x² + 3x + b
                            –3x⁴ – 6x³                  – 3ax
                          (+)  (+)                        (+)            
                            2x³ + (3 – a)x² + (3 + 3a)x + b
                            2x³ + 4x²                            + 2a
                         (–)   (–)                                 (–)     
                          –(1 + a)x² + (3 + 3a)x + (b – 2a)

⇒ –(1 + a)x² + (3 + 3a)x + (b – 2a) = 0

= 0.x² + 0.x + 0

On comparing the coefficient of x² and constant term, we get

a + 1 = 0

⇒ a = – 1

And b – 2a = 0

⇒ b = 2a

⇒ b = 2(–1) = –2

For a = –1 and b = –2,

The zeroes of q(x) are also the zeroes of the polynomial p(x).q(x) = x³ + 2x² – 1

And p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x – 2

Since, dividend = Divisor × Quotient + Remainder

p(x) = (x³ + 2x² – 1)(x² – 3x + 2) + 0

= (x³ + 2x² – 1)(x – 2)(x – 1)

So, the zeroes of p(x) are 1 and 2 which are not the zeroes of q(x).