*Example 2*Given that

*x*

^{2}+

*x*- 6 is a factor of 2

*x*

^{4}+

*x*

^{3}-

*ax*+

*bx*+

*a*+

*b*- 1, find the value of

*a*and

*b*.

*x*^{2} + * x* - 6 = (*x* + 3)(*x* + 2)

Let

*f*(*x*) = 2*x*^{4}+*x*^{3}-*ax*^{2}+*bx*+*a*+*b*- 1*f*(-3) = 2(-3)

^{4}+ (-3)

^{3}-

*a*(-3)

^{2}- 3

*b*+

*a*+

*b*- 1 = 0

134 - 8

*a*- 2*b*= 0 4

*a*+*b*= 67 --------(1)*f*(2) = 2(2)^{4} + 2^{3} - *a*(2)^{2} + 2*b* + *a* + * b* - 1 = 0

39 - 3

*a*+ 3*b*= 0*a*-

*b*= 13 --------(2)

(1) + (2) : 5*a* = 80

*a*= 16

when

*a*= 16,*b*= 3*Example 3
*

*k*for which (

*a*+ 2

*b*) where

*a*does not = 0 and

*b*does not = 0, is a factor of

*a*

^{4}+ 32

*b*

^{4}+

*a*

^{3}

*b*(

*k*+ 3).

Solution

Let

Let

*f*(*a*) =*a*^{4}+ 32*b*^{4}+*a*^{3}*b*(*k*+ 3)*f*(-2

*b*) = (-2

*b*)

^{4}+ 32

*b*

^{4}+ (-2

*b*)

^{3}

*b*(

*k*+ 3) = 0

48

*b*^{4}- 8*b*^{4}(*k*+ 3) = 0 8

*b*^{4}[6 - (*k*+ 3)] = 0 8

*b*^{4}(3 -*k*) = 0 Since

*b*does not = 0, 3 -*k*= 0 k = 3

*Example 4*Determine the value of

*k*for which

*x*+ 2 is a factor of (

*x*+ 1)

^{7}+ (2

*x*+

*k*)

^{3}.

Solution

Let f(x) = (

Let f(x) = (

*x*+ 1) + (2*x*+*k*)*f*(-2) = (-2 + 1) + (-4 +

*k*) = 0

(

*k*- 4) - 1 = 0 (

*k*- 4) = 1*k*- 4 = 1

*k*= 5

*Example 5*Given that

Hence (i) factorise

*f*(*x*) =*x*^{3}-*x*^{2}-*ax*-*b*, where*a*and*b*are constants, has the factor*x*- 3 but has a remainder of 13*x*- 11 when divided by*x*+ 4. Calculate the values of*a*and*b*.Hence (i) factorise

*f*(*x*) completely. (ii) Solve the equation 27*x*^{3}- 3*ax*=*ax*^{2 }+*b*.*Solution*

*f*(

*3*) = 0

3

^{3}- 3^{2}- 3*a*-*b*= 0 3

*a*+*b*= 18 --------(1)*f*(-4) = (-4)

^{3}- (-4)

^{2}-

*a*(-4) -

*b*= 13(-4) - 11

4

*a*-*b*= 17 --------(2) sub (1) into (2):

*a*= 5,*b*= 3 (i) Factorising.

*f*(

*x*) = (

*x*- 3)(

*x*

^{2}+ 2

*x*+ 1)

= (

*x*- 3)(*x*+ 1)^{2} (ii) 27

*x*^{3}- 9*x*^{2}- 3*ax*-*b*= 0 (3

*x*)^{3}- (3*x*)^{2}-*a*(3*x*) -*b*= 0 Let

*y*= 3*x**y*

^{3}-

*y*

^{2}-

*ay*-

*b*= 0

(

*y*- 3)(*y*+ 1)^{2}= 0 [from (1)] y = 3, y = -1

*x*= 1,

*x*= -1/3

*Example 6*

Show that the expression

Solution

*x*^{3}+ (k-2)*x*^{2}+*(k-7)x*- 4 has a factor x+1 for all values of k .If the expression also has a factor x+2, find the value of k and the third factor.Solution

Let f(x) =

*x*

^{3}+ (k-2)

*x*

^{2}+

*(k-7)x*- 4

f(-1) =(-1)

^{3}+ (k-2)(-1)

^{2}+

*(k-7)*(-1) - 4 = -1+ k - 2 - k + 7 - 4 = 0

Therefore (x+1) is a factor of f(x).

x + 2 is a factor of f(x). -------(given)

Therefore f(-2)=0

(-2)

^{3}+ (k-2)(-2)

^{2}+

*(k-7)*(-2) - 4 = 0

-8 + 4k -8 - 2k +14 - 4 = 0

2k = 6

k = 3

Therefore f(x) =

*x*

^{3}+

*x*

^{2}- 4

*x*- 4

Therefore (x + 1) (x + 2) = x

^{2}+ 3x + 2 is also a factor of f(x).

Therefore the third factor of f(x) is x - 2.

*Example 7*Given that kx

^{3}+ 2x

^{2}+ 2x + 3 is a factor of kx

^{3}- 2x + 9, have a common factor, what are the possible values of k?

Solution

Let f(x) = >kx

^{3}+ 2x

^{2}+ 2x + 3 and g(x) = kx

^{3}- 2x + 9.

Let (x - c) be a common factor of both f(x) and g(x).Therefore ...

f(c) = 0

kc

^{3}+ 2c

^{2}+ 2c + 3 = 0 ---------(1)

g(c) = 0

kc

^{3}- 2c + 9= 0 ---------(2)

Subtracting equation(2) from (1),we have

2c

^{2}+ 4c - 6 = 0

c

^{2}+ 2c - 3 = 0

(c + 3) (c - 1) =0

c = -3 or c = 1

Substituting c = -3 or c = 1 in equation (2), we get

k = 5/9 when c = -3 and

k = -7 when c = 1