Sagot :
Answer:
In this page, 'Solutions to algebra-III' we are discussing how to do the problems given in problems on algebra-III.
Solutions to algebra-III
1. Find the remainder using remainder theorem, when
(i) 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.
(ii) 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.
(iii) 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.
(iv) 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.
(v) 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.
(vi) 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(1) + 12(-1) -2(1) -18(-1) +14
= 8 -12 -2 +18 +14
= 26
The remainder is 26.
. (vii) 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) = (a)³ - a(a)² - 5(a) +2a
= a³ - a³ -5a +2a
= -3a
The remainder is -3a.
2. 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
3. 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
4. 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
5. 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 by (1) and (2)
2m - m = 25 - 20
m = 5.
Substituting m = 5 in p(2) we get the remainder.
= 20 -5 = 15.
The remainder is 15
Step-by-step explanation:
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