Exercise 3.2
Polynomial Division:
When one number is divided by another:
2477
15
The answer contains a quotient and a remainder:
165 remainder: 2
15 |2477
So, the quotient is '165' and the remainder is '2'
If '15' had been a factor of '2477', then the remainder
would have been zero.
Similarly, when one polynomial is divided by another:
4x² - 2x² - 1
2x - 3
The answer is also made up of a quotient and a remainder
(again, the remainder will be zero if the divisor is a factor of
the function).
But how do we perform such a division:
Step 1: Write out the division. Write
the terms out in degree order and include all terms of each
expression (even the zero terms) :
.
2x - 3 │ 4x³ - 2x² + 0x - 1
Step 2: Look at the first term of each
expression. In this case, ask what does '2x' have to be
multiplied by to get '4x3'? Write the answer (2x2)
on top of the division:
2x² .
2x - 3 │ 4x³ - 2x² + 0x - 1
Step 3: Now multiply the whole of the
divisor by '2x2'. Write the answer under the 1st two
terms:
2x² .
2x - 3 │ 4x³ - 2x² + 0x - 1
4x³ - 6x²
Step 4: Now subtract these new terms
from the terms above:
2x² .
2x - 3 │ 4x³ - 2x² + 0x - 1
— 4x³ - 6x²
4x²
Step 5: Bring down the remaining
terms:
2x² .
2x - 3 │ 4x³ - 2x² + 0x - 1
— 4x³ - 6x²
4x² + 0x - 1
Step 6: Now return to step 2 (i.e.
what do we have to multiply '2x' by to get '4x2'?...
2x² + 2x .
2x - 3 │ 4x³ - 2x² + 0x - 1
— 4x³ - 6x²
4x² + 0x - 1
And keep repeating these steps until you cannot do it any
more!
2x² + 2x + 3 .
2x - 3 │ 4x³ - 2x² + 0x - 1
— 4x³ - 6x²
4x² + 0x - 1
— 4x² - 6x
+ 6x - 1
— 6x - 9
8
Step 7: Since we can't divide any
more, the quotient is '2x2 + 2x +
6' and the remainder is '8'
The Remainder Theorem
If we just want the remainder when a polynomial is divided by
a linear function: e.g.
4x³ - 2x² - 1
2x - 3
Then, rather than performing the whole division:
.
2x - 3 │ 4x³ - 2x² + 0x - 1 etc...
There is a quicker way:
Instead of dividing by '2x-3' we can just put x = 3/2 into the
function. The answer gives the remainder:
f(3/2) = 4(3/2)3 - 2(3/2)2 - 1
= 4(27/8) - 2(9/4) - 1
= 8
This is the same answer as we got above, but much quicker!
Similarly:
Instead of dividing by 'x-1' we can put x=1 into the function
Instead of dividing by 'x+2' we can put x=-2 into the function
Instead of dividing by '2x-3' we can put x=3/2 into the function
Instead of dividing by 'ax+b' we can put x=-b/a into the
function
etc...
Note: The remainder theorem will not give us the quotient,
just the remainder
Note: We must use polynomial division for question 1 -
because we need to find the quotient
Question 1, part (a): Step 1: So the division is: .
x - 1 │ x³ + 2x² - x - 1
Step 2: Look at the first term of each
expression. In this case, ask what does 'x' have to be
multiplied by to get 'x3'? Write the answer (x2)
on top of the division:
x² .
x - 1 │ x³ + 2x² - x - 1
Step 3: Now multiply the whole of the
divisor by 'x2'. Write the answer under the 1st two
terms:
x² .
x - 1 │ x³ + 2x² - x - 1
x³ - x²
Step 4: Now subtract these new terms
from the terms above:
x² .
x - 1 │ x³ + 2x² - x - 1
— x³ - x²
3x²
Step 5: Bring down the remaining terms:
x² .
x - 1 │ x³ + 2x² - x - 1
— x³ - x²
3x² - x - 1
Step 6: Now return to step 2 (i.e. what
do we have to multiply 'x' by to get '3x2'?...
x² + 3x .
x - 1 │ x³ + 2x² - x - 1
— x³ - x²
3x² - x - 1
And keep repeating these steps until you cannot do it any
more!
x² + 3x + 2 .
x - 1 │ x³ + 2x² - x - 1
— x³ - x²
3x² - x - 1
— 3x² - 3x
+ 2x - 1
— 2x - 2
1
Step 7: Since we can't divide any more,
the quotient is 'x2 + 3x +
2' and the remainder is '1'
Note: We can use the remainder theorem
to check your answer: To find the remainder upon dividing by (x
- 1), we need to put in x = 1 into the equation... (If
the answer is 1,
then our division must be correct)
Question 1, part (b): Step 1: So the division is:
.
x + 1 │ x³ + 2x² - x - 1
Step 2: Look at the first term of each
expression. In this case, ask what does 'x' have to be
multiplied by to get 'x3'? Write the answer (x2)
on top of the division:
x² .
x + 1 │ x³ + 2x² - x - 1
Step 3: Now multiply the whole of the
divisor by 'x2'. Write the answer under the 1st two
terms:
x² .
x + 1 │ x³ + 2x² - x - 1
x³ + x²
Step 4: Now subtract these new terms
from the terms above:
x² .
x + 1 │ x³ + 2x² - x - 1
— x³ + x²
x²
Step 5: Bring down the remaining terms:
x² .
x + 1 │ x³ + 2x² - x - 1
— x³ + x²
x² - x - 1
Step 6: Now return to step 2 (i.e. what
do we have to multiply 'x' by to get 'x2'?...
x² + x .
x + 1 │ x³ + 2x² - x - 1
— x³ + x²
x² - x - 1
And keep repeating these steps until you cannot do it any
more!
x² + x - .. .
x + 1 │ x³ + 2x² - x - 1
— x³ + x²
x² - x - 1
— x² + x
- 2x - 1
— - 2x - 2
1
Step 7: Since we can't divide any more,
the quotient is '.......' and the remainder is
'1'
Note: We can use the remainder theorem
to check your answer: To find the remainder upon dividing by (x
+ 1), we need to put in x = -1 into the equation... (If
the answer is 1,
then our division must be correct)
Question 1, part (c): Step 2: Look
at the first term of each expression. In this case, ask what
does 'x' have to be multiplied by to get 'x3'? Write
the answer (x2) on top of the division:
x² .
x - 2 │ x³ + 2x² - x - 1
Step 3: Now multiply the whole of the
divisor by 'x2'. Write the answer under the 1st two
terms:
x² .
x - 2 │ x³ + 2x² - x - 1
x³ - 2x²
Step 4: Now subtract these new terms
from the terms above:
x² .
x - 2 │ x³ + 2x² - x - 1
— x³ - 2x²
4x²
Step 5: Bring down the remaining terms:
x² .
x - 2 │ x³ + 2x² - x - 1
— x³ - 2x²
4x² - x - 1
Step 6: Now return to step 2 (i.e. what
do we have to multiply 'x' by to get '4x2'?...
x² + 4x .
x - 2 │ x³ + 2x² - x - 1
— x³ - 2x²
4x² - x - 1
And keep repeating these steps until you cannot do it any
more!
x² + 4x + .. .
x - 2 │ x³ + 2x² - x - 1
— x³ - 2x²
4x² - x - 1
— 4x² - 8x
+ 7x - 1
— ..x - ..
..
Step 7: Since we can't divide any more,
the quotient is '.......' and the remainder is
'..'
Note: We can use the remainder theorem
to check your answer: To find the remainder upon dividing by (x
- 2), we need to put in x = 2 into the equation...
Question 1, part (e): Step 2: Look
at the first term of each expression. In this case, ask what
does '2x' have to be multiplied by to get 'x3'? Write
the answer (½x2) on top of the division:
½x² .
2x - 1 │ x³ + 2x² - x - 1
Step 3: Now multiply the whole of the
divisor by 'x2'. Write the answer under the 1st two
terms:
½x² .
2x - 1 │ x³ + 2x² - x - 1
x³ - ½x²
Step 4: Now subtract these new terms
from the terms above:
½x² .
2x - 1 │ x³ + 2x² - x - 1
— x³ - ½x²
2½x²
Step 5: Bring down the remaining terms:
½x² ..
2x - 1 │ x³ + 2x² - x - 1
— x³ - ½x²
2½x² - x - 1
Step 6: Now return to step 2 (i.e. what
do we have to multiply '2x' by to get '5/2 x2'?...
½x² + 1¼x .
2x - 1 │ x³ + 2x² - x - 1
— x³ - ½x²
2½x² - x - 1
And keep repeating these steps until you cannot do it any
more!
½x² + 1¼x + .. .
2x - 1 │ x³ + 2x² - x - 1
— x³ - ½x²
2½x² - x - 1
— 2½x² - 1¼x
+ ¼x - 1
— ¼x - ..
..
Step 7: Since we can't divide any more,
the quotient is '.......' and the remainder is
'..'
Note: We can use the remainder theorem
to check your answer: To find the remainder upon dividing by (2x
- 1), we need to put in x = ½ into the equation...
Question 1, part (h): So the division is:
.
x + t │ x³ + 2x² - x - 1
Step 2: Look at the first term of each
expression. In this case, ask what does 'x' have to be
multiplied by to get 'x3'? Write the answer (x2)
on top of the division:
x² .
x + t │ x³ + 2x² - x - 1
Step 3: Now multiply the whole of the
divisor by 'x2'. Write the answer under the 1st two
terms:
x² .
x + t │ x³ + 2x² - x - 1
x³ + tx²
Step 4: Now subtract these new terms
from the terms above:
x² .
x + t │ x³ + 2x² - x - 1
— x³ + tx²
(2-t)x²
Step 5: Bring down the remaining terms:
x² .
x + t │ x³ + 2x² - x - 1
— x³ + tx²
(2-t)x² - x - 1
Step 6: Now return to step 2 (i.e. what
do we have to multiply 'x' by to get '(2-t)x2'?...
x² + (2-t)x .
x + t │ x³ + 2x² - x - 1
— x³ + tx²
(2-t)x² - x - 1
And keep repeating these steps until you cannot do it any
more!
x² + (2-t)x - .. .
x + t │ x³ + 2x² - x - 1
— x³ + tx²
(2-t)x² - x - 1
— (2-t)x² + (2t-t²)x
etc...
Step 7: Since we can't divide any more,
the quotient is '.......' and the remainder is
'..'
From question 2 onwards, you can use the remainder theorem
alone...
Question 2, part (a): To find the remainder when x²
- 7x - 3 is divided by (x + 2), we just put x=-2 into the
equation
f(x) = x² - 7x - 3
f(-2) = (-2)² - 7(-2) - 3
= ...
Question 3, part (a): To find the remainder when '1 - 2x - 3x²'
is divided by 'x-2', we just put x=2 into the equation:
f(2) = 1 - 2(2) -
3(2)² = ...
Question 3, part (b): This time, we need to put x=½ into the
equation: f(½) = (½)² -
5(½) - 3 = ...
Question 3, part (c): This time, we put x=-3 into the
equation ...
Question 3, part (d): Just put x=... into the equation ...
Question 6: Again, we just put x=-1 into the equation
(the
answer will be in terms of a, b, c & d)
Question 8: Again, putting x=-2 into the equation gives the
remainder but, since we don’t know ‘a’, the answer will be
in terms of ‘a’ Since the question tells us the remainder is 0, we can make our
answer = 0 and solve to find ‘a’
Question 12: If we put in x=1, we get the remainder upon
dividing by (x-1) (the answer will be in terms of p and q)
Similarly, putting in x=-2 gives the remainder upon dividing
by (x+2) (again in terms of p and q)
Since the question says these remainders are equal, we can
make our answers equal and simplify to get 3p - q = -4 (call
this equation 1)
Finally, to get the remainder upon dividing by (x-2), we put
in x=2 into the equation. The question tells us we can make this
answer = -18 to get another equation (call this equation 2)
Solve equations 1 & 2 (simultaneously) to find p &
q...
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