Find the values of the following polynomials at the indicated values of the variables.
- (i) 5x2 − 3x + 7 if x = 1
- (ii) 4t3 − t2 + 6 if t = a
Answer: (i) 9 (ii) 4a3 − a2 + 6
Step-by-step solution
Idea: Replace the variable by the given value everywhere. If the value is another letter, the answer is an expression in that letter.
(i) 5x2 − 3x + 7 if x = 1
- 5(1)2 − 3(1) + 7 = 5 − 3 + 7½ mark
- = 9½ mark
9
(ii) 4t3 − t2 + 6 if t = a
- Put a in place of t: 4(a)3 − (a)2 + 6½ mark
- = 4a3 − a2 + 6. This is as far as it goes: a is not a number we know.½ mark
4a3 − a2 + 6
(i) 9 (ii) 4a3 − a2 + 6
Check: (ii) with a = 1 gives 4 − 1 + 6 = 9, and the original polynomial at t = 1 is also 9 ✓.
Answer to write in the exam
(i)
5(1)2 − 3(1) + 7 = 5 − 3 + 7
∴ Value = 9
(ii)
4(a)3 − (a)2 + 6
∴ Value = 4a3 − a2 + 6
Common mistakes that cost marks
- In (i), writing 5x2 at x = 1 as 25 (squaring the 5).
- In (ii), trying to combine 4a3 and −a2 into 3a5 or 3a. They are unlike terms.
- In (ii), writing 4a3 as (4a)3 = 64a3.
How this can come in the exam
MCQ (1 mark)
If p(x) = x2 − 2x, then p(k + 1) equals
- k2 − 1
- k2 − 2k
- k2 + 1
- k2 − 2
Show answer
(A) k2 − 1
(k + 1)2 − 2(k + 1) = k2 + 2k + 1 − 2k − 2 = k2 − 1.
Short answer (2 marks)
Find the value of 2m3 + 3m − 4 at m = −2 and at m = b.
Show answer
At −2: −16 − 6 − 4 = −26 (1 mark). At b: 2b3 + 3b − 4 (1 mark).Try one yourself
Find the value of 3x2 + x − 8 at x = 2 and at x = −c.
Show answer
At 2: 12 + 2 − 8 = 6. At −c: 3c2 − c − 8.
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