Question 29
For which values of a and b, the zeroes of q(x)=x3+2x2+a are also the zeroes of the polynomial p(x)=x5−x4−4x3+3x2+3x+b? Which zeores of p(x) are not the zeroes of q(x)?
Solution:
To solve the problem, we need to find the values of a and b such that the zeros of the polynomial q(x)=x3+2x2+a are also the zeros of the polynomial p(x)=x5−x4−4x3+3x2+3x+b. We will also determine which zeros of p(x) are not zeros of q(x).
1. Understanding the Relationship:
Since the zeros of q(x) are also zeros of p(x), it implies that q(x) is a factor of p(x). Therefore, when we divide p(x) by q(x), the remainder should be zero.
2. Perform Polynomial Long Division:
We will divide p(x) by q(x) using polynomial long division.
– Setup the Division:
p(x)=x5−x4−4x3+3x2+3x+b
q(x)=x3+2x2+a
– First Division:
Divide the leading term of p(x) by the leading term of q(x):
x5/x3=x2
Multiply q(x) by x2:
x2(x3+2x2+a)=x5+2x4+ax2
Subtract this from p(x):
(x5−x4−4x3+3x2+3x+b)−(x5+2x4+ax2)=−3x4−4x3+(3−a)x2+3x+b
– Second Division:
Divide the leading term of the new polynomial by the leading term of q(x):
−3x4/x3=−3x
Multiply q(x) by −3x:
−3x(x3+2x2+a)=−3x4−6x3−3ax
Subtract:
(−3x4−4x3+(3−a)x2+3x+b)−(3x4−6x3−3ax)=2x3+(3−a)x2+(3+3a)x+b
– Third Division:
Divide the leading term again:
2x3/x3=2
Multiply q(x) by 2:
2(x3+2x2+a)=2x3+4x2+2a
Subtract:
(2x3+(3−a)x2+(3+3a)x+b)−(2x3+4x2+2a)=(3−a−4)x2+(3+3a)x+(b−2a)
3. Setting the Remainder to Zero:
For q(x) to be a factor of p(x), the remainder must be zero:
(3−a−4)x2+(3+3a)x+(b−2a)=0
This implies:
– Coefficient of x2: 3−a−4=0 → a=−1
– Coefficient of x: 3+3a=0 → 3−3=0 (consistent)
– Constant term: b−2a=0 → b−2(−1)=0 → b=−2
4. Final Values:
The values of a and b are:
a=−1,b=−2
5. Finding Extra Zeros of p(x):
Now, we need to find the zeros of p(x) that are not zeros of q(x). The remainder polynomial after division was:
x2−3x+2=0
Factoring gives:
(x−1)(x−2)=0
Thus, the zeros are x=1 and x=2.
Conclusion:
The zeros of p(x) that are not zeros of q(x) are x=1 and x=2.