## Solutions to algebra-II

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

### Solutions to algebra-II

1.       Find the zeros of the following polynomials

(i)     p(x) = 4x-1

Solution:  Given p(x)    = 4x-1

= 4(x-1/4).

we have p(1/4) = 4(1/4-1/4)

= 0.

Hence x= 1/4 is the zero of the polynomial.

(ii)    p(x) = 3x+5

Solution:  Given p(x)  = 3x+5

=  3(x+5/3)

p(-5/3) =  3(-5/3)+5

=    -5   +  5

=     0.

Hence x= -5/3 is the zero of the polynomial.

(iii)   p(x) =  2x

Solution:  Here p(0)  = 2(0) =0

Hence x=0 is the zero of the polynomial.

(iv)   p(x) =  x+9

Solution:   Given p(x) = x+9

p(-9) = -9+9 =0

Hence x = -9 is the zero of the polynomial.

2.        Find the roots of the following polynomial equations.

(i)    x-3   = 0

Solution:  Given x-3 =0

Which implies that x =3.

∴ x =3 is a root of x-3 = 0

(ii)   5x-6  = 0

Solution: Given 5x-6 =0.

Which implies that 5x = 6

x  = 6/5

∴  x = 6/5 is a root of 5x-6 = 0.

(iii)  11x +1 = 0

Solution:  Given 11x + 1 =0

Which implies that 11x  = -1

x  = -1/11

∴ x = -1/11 is a root of 11x+ 1=0.

(iv)  -9x      = 0

Solution:    Given that -9x =0

Which implies that        x = 0/9

x = 0

∴  x =0 is a root of the equation -9x = 0

3.       Verify whether the following are roots of the polynomial equations           indicated against them.

(i)   x² - 5x +6 = 0:      x= 2,3

Solution:

(a)   Given that p(x) =  x² - 5x +6

p(2) =  2² 5(2) +6

=  4 -10 + 6

= 0

∴   x = 2 is the root of the equation  x² - 5x +6

(b)  Given that p(x)   =  x² - 5x +6

p(3)  =  3² 5(3) +6

=  9 -15 + 6

= 0

∴   x = 3 is the root of the equation  x² - 5x +6

(ii)  x² +4x+3  = 0:      x= -1,2

Solution:

(a)   Given that p(x) =  x² +4x +3

p(-1) =  (-1)²+ 4(-1) +3

=   1  -4 + 3

=    0

∴   x = -1 is the root of the equation  x² + 4x +3

(b)  Given that p(x)   =  x² + 4x +3

p(2)  =  2²  +4(2) +3

=  4  + 8 + 3

≠    0

∴   x = 2 is not the root of the equation  x² +4x +3

(iii) x³ - 2x² -5x +6 = 0:    x = 1, -2,3

Solution:

(a)   Given that p(x) =  x³ - 2x² -5x +6

p(1) =  (1)³ -2(1)²-5(1) +6

=   1  -2-5 + 6

=    0

∴   x = 1 is the root of the equation   x³ - 2x² -5x +6

(b)  Given that p(x)   =   x³ - 2x² -5x +6

p(-2)  =  (-2)³  -2(-2)² -5(-2)+6

=  -8 - 8 +10 +6

=    0

∴   x = -2 is  the root of the equation  x³ - 2x² -5x +6

(c)    Given that p(x)   =   x³ - 2x² -5x +6

p(3)  =  (3)³  -2(3)² -5(3)+6

=  27   -18   -15  +6

=   0

∴   x = 3 is not the root of the equation  x³ - 2x² -5x +6

(iv)  x³ - 2x² -x +2 = 0:     x =  -1,2,3

Solution:

(a)   Given that p(x) =  x³ - 2x² -x +2

p(-1) =  (-1)³ -2(-1)²-(-1) +2

=   -1  -2+1 + 2

=    0

∴   x = -1 is the root of the equation   x³ - 2x² -x +2

(b)  Given that p(x)   =   x³ - 2x² -x +2

p(2)  =  (2)³  -2(2)² -(2)+2

=  8 - 8 -2 +2

=    0

∴   x = 2 is  the root of the equation  x³ - 2x² -x +2

(c)    Given that p(x)   =   x³ - 2x² -x +2

p(3)  =  (3)³  -2(3)² -(3)+2

=  27   -18   -3  +2

≠    0

∴   x = 3 is not the root of the equation  x³ - 2x² -x +2

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