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MAT1511 EXAM PACK 2022
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MAT1511EXAM PACK
FOR ASSISTANCE WITH THIS MODULE +27 67 171 1739
, ONLY FOR SEMESTER 1 STUDENTS
ASSIGNMENT 01
Submission date: 1 March 2019
Note: All references are for the 6th edition of Steward, Redlin and Waston.
Question 1
Let P (x) = x6 − 2x5 − x4 + x3 + 2x2 + x − 2
1
1(a) Determine whether x + 2
is a factor of P (x) , use Factor Theorem.
Solution:
Concept Explained:
See Factor Theorem on page 249.
Let P (x) = (x − c) . Q (x) + r
where Q (x) is the quotient and r is the remainder when P (x) is divided by (x − c) .
Set x − c = 0, we have x = c. Thus the Factor Theorem implies that if P (c) = 0,
then x − c is a factor of P (x) .
1 1
In this question, x+ represents (x − c) . Hence we need to test if P − = 0.
2 2
6 5 5 4 3 2
1 1 1 1 1 1 1 1
P − = − −2 − − − − − + − +2 − + − −2
2 2 2 2 2 2 2 2
159
=− 6= 0
64
1
Hence, x + is not a factor of P (x) .
2
1(b) Find all the possible rational zeros of P (x) by using the Rational Zeros Theorem
Concept Explained: (page 253)
The Rational Zeros Theorem states that all the possible rational zeros of a P (x) is
p
given by , where p is the factor of the constant coefficient and q is the factor of the
q
leading coefficient
2
, Solution:
Since P (x) = x6 − 2x5 − x4 + x3 + 2x2 + x − 2
p is ± 1 or ± 2
p
Therefore, the possible values of are:
q
±1 ±2
and
±1 ±1
These give ±1 and ±2. That is
−1, −2 + 1, and + 2
1(c) P (x) = x6 − 2x5 − x4 + x3 + 2x2 + x − 2
Solve P (x) = 0
Concept Explained:
To solve P (x) = 0, first Rational Zero Theorem 1(b) may be used to find all possible
roots/zeros of P (x) , secondly use Factor Theorem (1a) to determine the actual roots/zeros
of P (x) . After complete factorisation, P (x) = 0 becomes A · B · C · D... = 0 where each
A, B, C, D represents factor of P (x) . Lastly set each of these to zero as follows
A = 0 or B = 0 or C = 0, or D = 0 ...
Then each of these give respective values of x.
Solution:
From the Rational Zero Theorem in 1(b), the possible roots/zeros of P (x) are −1, −2, +1 and
+2. Factor Theorem is applied to determine the actual roots as follows:
P (−1) = (−1)6 − 2 (−1)5 − (−1)4 + (−1)3 + 2 (−1)2 + (−1) − 2
1+2−1−1+2−1−2=0
P (−2) = (−2)6 − 2 (−2)5 − (−2)4 + (−2)3 + 2 (−2)2 + (−2) − 2
= 26 + 2 (25 ) − (24 ) − (23 ) + 2 (22 ) − 2 − 2
= 108 6= 0
P (1) = (1)6 − 2 (1)5 − (1)4 + (1)3 + +2 (1)2 + (1) − 2
=1−2−1+1+2+1−2=0
and
P (2) = 26 − 2 (25 ) − 24 + 23 + 2 (22 + 2 − 2)
= 64 − 64 − 16 + 8 + 8 + 2 − 2
=0
3
, Therefore (x + 1) , (x − 2) and (x − 1) are the roots of P (x) . But the highest power of P (x)
is 6, hence the roots are expected to be 6. Multiply the factors (x + 1) (x − 2) (x − 1) to have
x3 − 2x2 − x + 2. Use long division to divide P (x) by x3 − 2x2 − x + 2 to have x3 − 1. Thus
P (x) = x3 − 2x2 − x + 2 x3 − 1
Use Rational Zero Theorems to simplify x3 − 1 to have (x − 1) (x2 + x + 1) .
Therefore
P (x) = x3 − 2x2 − x + 2 (x − 1) x2 + x + 1
remember that x3 − 2x2 − x + 2 = (x + 1) (x − 2) (x − 1)
Hence
P (x) = (x + 1) (x − 2) (x − 1) (x − 1) (x2 + x + 1)
= (x − 1)2 (x + 1) (x − 2) (x2 + x + 1)
∴ P (x) = 0 implies
(x − 1)2 (x + 1) (x − 2) (x2 + x + 1) = 0.
⇒ x = −1, 1(twice), 2, and for x2 + x + 1 = 0
Use quadrafic formula √
−b ± b2 − 4ac
x= to get
2a
√ √
1 − 3 1 − 3
x= + i or + i
2 2 2 2
Sythetic division may also be used to determine the actual factors of P (x)
√ √
1 3 1 3
Therefore, the values of x for which P (x) = 0 are: −1, 1(twice), 2, − + i and − − i.
2 2 2 2
Question 2
Use Descartes’ Rule of Signs to determine the possible number of positive, negative and imaginary
zeros of
P (x) = 2x6 − 3x5 − 9x4 + 15x3 + 3x2 − 12x + 4.
Summarize your answer in the form of a table, see page 13 of the Study Guide.
4
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