If a polynomial f(x) is divided by (x-a), the remainder is f(a).
f(x) = (x-a) Q(x) + f(a)
How to find the remainder, when we divide a polynomial by linear.
Step 1 :
Equate the divisor to 0 and find the zero.
Step 2 :
Let p(x) be the given polynomial.
Step 3 :
Apply the zero in the polynomial to find the remainder.
Find the remainder using remainder theorem, when
Example 1 :
3x3+4x2-5x+8 is divided by x-1.
Solution :
Let p(x) = 3x3+4x2-5x+8. The zero of x-1 is 1.
When p(x) is divided by x-1, the remainder is p(1).
p(1) = 3(1)3+4(1)2-5(1)+8
= 3+4-5+8
= 10
The remainder is 10.
Example 2 :
5x3+2x2-6x+12 is divided by x+2.
Solution :
Let p(x) = 5x3+2x2-6x+12
The zero of x+2 is -2.
When p(x) is divided by x+2, the remainder is p(-2).
p(-2) = 5(-2)3+2(-2)2-6(-2)+12
= 5(-8)+2(4)+12+12
= 40 + 8 + 12 + 12
= -8
The remainder is -8.
Example 3 :
2x3-4x2+7x+6 is divided by x-2.
Solution :
Let p(x) = 2x3-4x2+7x+6
The zero of x-2 is 2.
When p(x) is divided by x-2, the remainder is p(2).
p(2) = 2(2)3-4(2)2+7(2)+6
= 2(8)-4(4)+14+6
= 16-16+14+6
= 20
The remainder is 20.
Example 4 :
4x3-3x2+2x-4 is divided by x+3
Solution :
Let p(x) = 4x3-3x2+2x-4. The zero of x+3 is -3.
When p(x) is divided by x+3, the remainder is p(-3).
p(-3) = 4(-3)3-3(-3)2+2(-3)-4
= 4(-27)-3(9)-6-4
= -108 -27 -6-4
= -145
The remainder is -145.
Example 5 :
4x3 - 12x2 +11x -5 is divided by 2x-1
Solution :
Let p(x) = 4x3 - 12x2 +11x -5. The zero of 2x-1 is 1/2.
When p(x) is divided by 2x-1, the remainder is p(1/2).
p(1/2) = 4(1/2)3-12(1/2)2 +11(1/2) -5
= 4(1/8) -12(1/4)+11/2-5
= -2
The remainder is -2.
Example 6 :
8x4+12x3-2x2-18x +14 is divided by x+1
Solution :
Let p(x) = 8x4+12x3-2x2-18x +14
The zero of x+1 is -1.
When p(x) is divided by x+1, the remainder is p(-1).
p(-1) = 8(-1)4+12(-1)3-2(-1)2-18(-1) +14
= 8-12-2+18+14
= 40-14
= 26
The remainder is 26.
Example 7 :
x3-ax2 -5x+2a is divided by x-a.
Solution :
Let p(x) = x3-ax2 -5x+2a. The zero of x-a is a.
When p(x) is divided by x-a, the remainder is a.
p(a) = a3-a(a)2 -5a+2a
= a3-a3 -3a
= -3a
The remainder is -3a.
Example 8 :
When the polynomial
2x3-ax2+9x-8
is divided by x-3 the remainder is 28. Find the value of a.
Solution :
Let p(x) = 2x3-ax2+9x-8
When p(x) is divided by x-3, the remainder is p(3).
Given that p(3) = 28.
This implies that 2(3)³ - a(3)² + 9(3) -8. = 28
2(27)-a(9)+27-8 = 28
54-9a+19 = 28
73-9a = 28
73-28 = 9a
45 = 9a
a = 5
So, the value of a is 5.
Example 9 :
Find the value of m if
x3-6x2+mx+60
leaves the remainder 2 when divided by (x+2).
Solution :
Let p(x) = x3-6x2+mx+60
When p(x) is divided by (x+2) the remainder is p(-2).
Given that p(-2) = 2
This implies that (-2)3-6(-2)2+m(-2)+60 = 2
-8-6(4)-2m+60 = 2
-8-24-2m+60 = 2
28-2m = 2
28-2 = 2m
26 = 2m
m = 13
Example 10 :
If (x-1) divides
mx3-2x2+25x-26
without remainder find the value of m.
Solution :
Let p(x) = mx3-2x2+25x-26
When p(x) is divided by (x-1), the remainder is p(1).
Given that p(1) = 0
This implies that m(1)3-2(1)2+25(1)-26 = 0
m-2+25-26 = 0
m-3 = 0
m = 3
Example 11 :
If the polynomials
x3+3x2-m and 2x3-mx+9
leave the same remainder when they are divided by (x-2), find the value of m. Also find the remainder.
Solution :
Let p(x) = x3+3x2-m
q(x) = 2x3-mx+9
When p(x) is divided by (x-2) the remainder is p(2). Now,
p(2) = 23+3(2)2-m
= 8+12-m
= 20 - m --------(1)
When q(x) is divided by (x-2) the remainder is q(2). Now,
q(2) = 2(2)3-m(2)+9
= 16-2m+9
= 25-2m --------(2)
Given that p(2) = q(2). That is
20 - m = 25 - 2m
2m - m = 25 - 20
m = 5
By applying the value of m in (1), we get
= 20 -5
= 15.
The remainder is 15.
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