OPERATION OF POLYNOMIAL FUNCTION
ADDITIONAL AND SUBTRACTION
The sum of two polynomials is found by adding the coefficients of terms of the same degree or like terms, and subtraction can be found by subtracting the coefficient of like terms.
EXAMPLE
EXAMPLE
If p(x) = 4x2– 3x+7, Q(x) =3x+2 and r(x)= 5x3-7x2+9
Solution
Sum =p(x)+q(x)+r(x)
(4x2-3x+7)+(3x+2)+(5x3-7x2+9)
4x2-3x+7+3x+2+5x3-7+9
4x2-3x+7+3x+2+5x3-7+9
4x2+5x2+7x3+3x+7+2+9
5x3-3x2+18
Alternatively the sum can be obtained by arranging them vertically
4x2-3x+7
0x2+3x+2
5x3-7x2+0+9
5x3-3x2+0+18
P(x)+q(x)+r(x)=5x3-3x2+18
Subtract -5x2+9+5 from 3x2+7x-2
Solution
(3x2+7x-2)-(-5x2+9x+5)
3x2+7x-2+5x2-9x-5
3x2+5x2+7x-9x-2-5
8x2-2x-7 answer
MULTIPLICATION
The polynomial R(x) and S(x) can be multiplied by forming all the product of terms from R(x) and terms from S(x) and then summing all the products by collecting the like terms.
The product can be denoted by RS(X)
The polynomial R(x) and S(x) can be multiplied by forming all the product of terms from R(x) and terms from S(x) and then summing all the products by collecting the like terms.
The product can be denoted by RS(X)
example
If p(x)=2x2-x+3 and q(x)=3x3-x find the product p(x)q(x)
Solution
P(x) q(x)= (2x2-x+3) (3x3-x)
6x5-2x3-3x4+x2+9x3-3x
6x5-3x4+7x3+x2-3x
DIVISION
The method used in dividing one polynomial by another polynomial of equal or lower degree is the same to the one used for the long division of number
Example:
1. Given p[x] =x3-3x2+4x+2 and q[x]=x-1 find
The method used in dividing one polynomial by another polynomial of equal or lower degree is the same to the one used for the long division of number
Example:
1. Given p[x] =x3-3x2+4x+2 and q[x]=x-1 find
Solution
x3-x2
0-2x2+4x
-2x2+2x
0+2x+2
2x-2
0+4
Therefore
= x2 – 2x + 2 + 
This means
X3– 3x2+4x+2 is dividend x-1 is divisor
X2-2x+2 is the quotient and 4 is the remainder
Then
X3-3x2+4x+2 = (x2-2x+2) (x-1) + 4
Divided quotient x divisor + remainder
2. Divide p(x) = x3-8 by q(x) = x-2
Solution:
-x3 -2x2
2x2-8
-2x2-4x
4x-8
4x-8
– –
– –
In the example [2] above there is a remainder i.e. the remainder is zero so dividing one function by another function is one of the way of finding the factors of the polynomial thus if we divide p(x) by q(x) is one of the factors of p(x) other factors can be obtained by fractionazing the quotient q(x).
3. Given p(x) = x3-7x+6 and q(x) = x+3 determine whether or not d(x) is one of the factors of p(x) and hence find the factors if p(x)
Solution
X3+3x2
-3x2-7x
-3x2-9x
2x+6
-2x+6
– –
Since the remainder is zero d(x) =x+3 is one of the factors of p(x)
To factorize
X2-3x+2
(x2-x) – (2x+2)
x(x-1)-2(x-1)
(x-2) (x-1)
Therefore other factors are (x-2) and (x-1)
THE REMAINDER THEOREM
The remainder theorem is the method of finding the remainder without using long division
Example
1. If p(x) = (x-2) and q(x)+8. Dividend divisor quotient remainder, Then by taking x-2=0 we find
x-2= 0
x=2
x-2= 0
x=2
subtracting x=2 in p(x)
p(2) =2-2 q(x) +r
p(2)= 0 x q(x)+r
p(2)= 0+r
p(2) =r
So the remainder r is the value of the polynomial p(x) when x=2
2. Give p(x) = x3-3x2+6x+5 is divided by d(x) = x-2 find the remainder using the remainder theorem
Solution
Let d(x)=0
x-2=0
X-2+2 =0+2
X=2
Subtracting x=2 in p(x)
P(2)= 23-3[2]2+6[2]+5
P(2)=8-12+12+5
P(2)=8+5
P(2)=13
The remainder is 13
3. The remainder theorem states that if the polynomial p[x] is divided by [x-a] then the remainder ‘r’ is given by p[a]
P(x) =(x-a) q(x)+r hence
P(a) =(a-a) q(x)+r
P(a)=o(q(x))+r
P(a)=0+r
P(a)=r
More Examples
1. By using the remainder theorem, Find the remainder when p[x] =4x2-6x+5 is divided by d[x] =2x-1
Solution
d (x)=0
2x-1=0
2x-1+1=0+1
X =
Substituting
X= in p(x)
in p(x) P() =4 x -6()+5
) =4 x 
-6()+5P()=1 – 6 x +5
)=1 – 6 x +5P()=1 -3+5
)=1 -3+5P()= 3
)= 3The remainder is 3
2. P(x)= 3x2-5x+5 is divided by d(x) =x+4
Solution
Let d(x)=0
X+4=0
X+4=0-4
X=-4
Subtracting x=-4 in p(x)
P(-4)=3(-4)2-=5(-4)+5
P(-4)=48+20+5
P(-4)=68+5
P(-4)= 73
The remainder is 73
3. P(x)= x3+2x2-x+4 is divided by d(x) = x+3
Solution
Let d(x)=0
X+3=0-3
X=-3
Substituting x=-3 in p(x)
P(-3)= -33+2(-3)2-(-3)+4
P(-3)=-27+18+3+4
P(-3)=-27+21+4
P(-3)=-27+25
P(-3)=-2
The remainder is -2
4. Find the value of ‘a’ if x3-3x2+ax+5 has the remainder of 17 when divided by x-3
Solution
By using remainder theorem
Let x-3 =0
x-3=0+3
x=3
Substituting x=3 we have
x3-3x2+ax+5=17
(3)3-3(3)2+a(3)+5=17
27-27+3a+5=17
3a+5=17-5
3a=12
a=4
5. If ax2+3x-5 has a remainder -3 when divided by x-2. Find the value of a.
Solution
By using the remainder theorem
Let x-2=0
x-2=0+2
x=2
subtracting x=2 we have
a(2)2+3(2)-5 =-3
4a+6-5= -3
4a+1= -3
4a= -4
a = -1
Exercise
1. Divide p(x) by d(x) in the following
P(x)=2x2+3x+7 d(x)=x2+4
P(x)=2x2+3x+7 d(x)=x2+4
Solution
Use remainder theorem to find the remainder when;
1. P(x)=x3-2x2+5x-4 is divide by d(x)=x-2
Solution
Let d[x]=0
x-2 + 2=0+2
x=2
substituting x=2 in p(x)
p(2)=23-2[2]2+5[2]-4
p(2)=8-8+10-4
p(2)=10-4
p(2)=10-4
p(2)=10-4
Therefore the remainder is 6
2. p(x)= 2x4+ x3+x- is divided by d(x)=x+2
is divided by d(x)=x+2solution
let d(x)=0
x+2 – 2 = 0-2
x = -2
Substituting x = – 2 in p(x)
p(-2)=2(-2)4+(-2)3+(-2)( –)
)p(-2)= 2(16)+(-8)-2( –)
)p(-2)=32-8-2 ( –)
)p(-2)=24-2( –)
)p(-2)=21
Therefore the remainder is 21
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