For which values of a & b are the zeroes of q(x)=x to power 3+2x power 2+a also the zeroes of the polymomials,p(x)=x power 5-x power 4-4x power 3+3x power 2+3x+b? Which zeroes of p(x) are not the zeroes of q(x)?
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Divide p(x) = x⁵ - x⁴ - 4 x³ + 3x² + 3x + b
by q(x) = x³ + 2x² + a
If the zeroes of q(x) are also zeroes of p(x) then the reminder should be 0.
x³ +2x² + a) x⁵ - x⁴ - 4x³ + 3x² + b ( x² - 3x + 2
x⁵ + 2x⁴ +ax²
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-3x⁴ + (3-a)x²- 4x³
-3x⁴ -3ax- 6x³
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2x³ +(3-a)x² + 3x(1+a) +(b-2a)
2x³ + 4x² + 2a
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- (a+1) x² + 3 (a+1) x +(b - 2a)
so a = -1 and b = +2 for the reminder to be 0.
the quotient x² - 3 x + 2 = (x - 2) (x - 1)
Hence the zeroes of p(x) are 1 and 2, which are in addition to the three zeroes of q(x).
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we can also find the zeroes of q(x) now.
q(x) = x³ + 2x² - 1
on examining the coefficients, x = -1 is a root.
q(x) = (x+1) (x² + x - 1)
x = -1 or [-1 + √5] / 2 or [-1 - √5]/2
by q(x) = x³ + 2x² + a
If the zeroes of q(x) are also zeroes of p(x) then the reminder should be 0.
x³ +2x² + a) x⁵ - x⁴ - 4x³ + 3x² + b ( x² - 3x + 2
x⁵ + 2x⁴ +ax²
===============
-3x⁴ + (3-a)x²- 4x³
-3x⁴ -3ax- 6x³
===================
2x³ +(3-a)x² + 3x(1+a) +(b-2a)
2x³ + 4x² + 2a
========================
- (a+1) x² + 3 (a+1) x +(b - 2a)
so a = -1 and b = +2 for the reminder to be 0.
the quotient x² - 3 x + 2 = (x - 2) (x - 1)
Hence the zeroes of p(x) are 1 and 2, which are in addition to the three zeroes of q(x).
===============
we can also find the zeroes of q(x) now.
q(x) = x³ + 2x² - 1
on examining the coefficients, x = -1 is a root.
q(x) = (x+1) (x² + x - 1)
x = -1 or [-1 + √5] / 2 or [-1 - √5]/2
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