## Solutions to algebra-III

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)       8x  + 12x³ -2x² - 18x +14 is divided by x+1

Solution:

Let p(x)  =   8x  + 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)  + 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.

Students can solve the problems on their own, compare the answer with the solutions discussed above in'Solutions to algebra-III'. If you are having any doubt you can contact us through mail, we will help you to clear your doubts.