## Solutions to algebra-IV

In this page, 'Solutions to algebra-IV' we are discussing how to do the problems given in problems on algebra-IV.

### Solutions to algebra-IV

1.          Determine whether (x+1) is a factor of the following polynomials.

(i)      6x + 7x³ - 5x - 4

Solution:

By factor theorem, if p(-1) = 0, then (x+1) is a factor of                       p(x)   =    6x + 7x³ - 5x - 4.

p(-1)  =    6(-1) + 7(-1)³ - 5(-1) - 4

=      6    -    7   +  5  - 4

=       0

∴  (x+1) is the factor of 6x + 7x³ - 5x - 4.

(ii)     2x + 9x³ + 2x² + 10x + 15

Solution:

By factor theorem, if p(-1) = 0, then (x+1) is a factor of

p(x)     =   2x + 9x³ + 2x² + 10x + 15

p(-1)    =   2(-1) + 9(-1)³ + 2(-1)² + 10(-1) + 15

=     2    -    9    +   2  -  10  +  15

=     0

∴ (x+1) is the factor of 2x + 9x³ + 2x² + 10x + 15

(iii)    3x³ + 8x² - 6x - 5

Solution:

By factor theorem, if p(-1) = 0, then (x+1) is a factor of

p(x)     =    3x³ + 8x² - 6x - 5

p(-1)    =    3(-1)³  + 8(-1)² - 6(-1) - 5

=      -3  +  8  +  6  - 5

≠       0

(x+1) ix not the factor of   3x³ + 8x² - 6x - 5

(iv)     x³ -  14x² + 3x + 12

Solution:

By factor theorem, if p(-1) = 0, then (x+1) is a factor of

p(x)     =    x³ -  14x² + 3x + 12

p(-1)    =   (-1)³ - 14(-1)²  + 3(-1) + 12

=     -1   - 14   -3 +12

≠     0

∴(x+1) is not the factor of x³ -  14x² + 3x + 12

2.          Determine whether (x+4) is a factor of x³ + 3x² - 5x + 36.

Solution:

By factor theorem, if p(-4) = 0, then (x+4) is a factor of

p(x)      =     x³ + 3x² - 5x + 36.

p(-4)     =    (-4)³ + 3(-4)² - 5(-4) + 36

=     -64 +  48 + 20 + 36

≠         0

∴(x+4) is not the factor of x³ + 3x² - 5x + 36

3.          Using factor theorem show that (x-1) is a factor of

4x³ - 6x² + 9x - 7.

Solution:

By factor theorem, if p(1) = 0, then (x-1) is a factor of

p(x)        =    4x³ - 6x² + 9x - 7

p(1)        =    4(1)³ - 6(1)² + 9(1) - 7

=     4 - 6 + 9 - 7

=      0

∴ (x-1) is the factor of 4x³ - 6x² + 9x - 7.

4.         Determine whether (2x+1) is a factor of 4x³ + 4x² - x - 1.

Solution:

By factor theorem, if p(-1/2) = 0, then (2x+1) is a

factor of  p(x)          =    4x³ + 4x² - x - 1

p(-1/2)      =     4(-1/2)³ + 4(-1/2)² - (-1/2) - 1

=     -1/2 + 1 + 1/2 - 1

=     0

∴ (2x+1) is the factor of 4x³ + 4x² - x - 1.

5.         Determine the value of p if (x+3) is a factor of x³ - 3x² - px + 24.

Solution:

By factor theorem, if p(-3) = 0, then (x+3) is a

factor of p(x)           =  x³ - 3x² - px + 24  =  0

p(-3)          =  (-3)³ - 3(-3)² - p(-3) + 24 = 0

This implies that  -27 - 27 + 3p + 24 = 0

-30 + 3p    =  0

3p    =  30

p    =  10

∴ The value of p = 10.

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