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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)?
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Solution
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).
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