Show that x=#1/4# is one of the roots of equation 4#x^3#-#x^2#-4x+1=0 Factorise 4#x^3#-#x^2#-4x+1 completely. Hence,solve (pls see below).?

(a)4#x^6#-#x^4#-4#x^2#+1=0
(b)#x^3#-4#x^2#-x+4=0

2 Answers
Narad T.
Jul 2, 2018

Please see the explanation below.

Explanation:

Let

#f(x)=4x^3-x^2-4x+1#

Then,

#f(1/4)=4*1/4^3-1/4^2-4/4+1=1/13-1/13-1+1=0#

Therefore,

#x=1/4# is a root of #f(x)#

You can perform a long division

#(x^3-x^2/4-x+1/4)/(x-1/4)=x^2-1=(x+1)(x-1)#

Similarly,

#f(1)=4-1-4+1=0#

#x=1# is a root of #f(x)#

#f()-1=-4-1+4+1=0#

#x=-1# is a root of #f(x)#

Therefore,

#4x^3-x^2-4x+1=4(x-1/4)(x-1)(x+1)#

graph{4x^3-x^2-4x+1 [-4.93, 4.934, -2.465, 2.465]}

Let

#g(x)=4x^6-x^4-4x^2+1#

#g(1)=4-1-4+1=0#

#g(-1)=4-1+4+1=0#

#g(1/2)=4*1/2^6-1/2^4-4*1/2^2+1=0#

#g(-1/2)=4*1/2^6-1/2^4-4*1/2^2+1=0#

#4x^6-x^4-4x^2+1=4(x-1)(x+1)(x-1/2)(x+1/2)#

graph{4x^6-x^4-4x^2+1 [-3.462, 3.467, -1.732, 1.73]}

Let

#h(x)=x^3-4x^2-x+4#

#h(1)=1-4-1+4=0#

#h(-1)=-1-4+1+4=0#

#h(4)=64-64-4+4=0#

Therefore,

#x^3-4x^2-x+4=(x-1)(x+1)(x-4)#

graph{x^3-4x^2-x+4 [-18.02, 18.01, -9.01, 9.01]}

Ratnaker Mehta
Jul 2, 2018

# 4x^3-x^2-4x+1=(4x-1)(x-1)(x+1)#.

#(a):"The real roots of "4x^6-x^4-4x^2+1," are, "+-1/2,+-1#.

#"The complex roots are, "+-1/2,+-1,+-i#.

Explanation:

If we plug in #x=1/4# in the given eqn., we have,

#"The L.H.S."=4(1/4)^3-(1/4)^2-4(1/4)+1#,

#=4(1/64)-(1/16)-1+1#,

#=1/16-1/16-1+1#,

#=0#,

#="The R.H.S."#

Hence, the Proof.

In other words, this also means that #(4x-1)# is a factor of the

poly. #p(x)=4x^3-x^2-4x+1#.

#"Now, "p(x)=4x^3-x^2-4x+1#,

#=x^2(4x-1)-1(4x-1)#,

#=(4x-1)(x^2-1)#.

# p(x)=4x^3-x^2-4x+1=(4x-1)(x-1)(x+1)#.

Hence, the factorisation.

Part (a) :

We observe that,

#4x^6-x^4-4x^2+1=4(x^2)^3-(x^2)^2-4(x^2)+1#.

So, if we let, #x^2=t," the eqn. "4x^6-x^4-4x^2+1=0#,

becomes, #4t^3-t^2-4t+1=0, i.e., p(t)=0#.

Knowing that, the zeroes of #p(x)# are, #x=1/4,+-1#,

the zeroes of #p(t)# have to be, #t=x^2=1/4,+-1#.

Clearly, the real zeroes of #4x^6-x^4-4x^2+1# are,

#+-1/2,+-1#.

N.B. :- The complex zeroes are, #+-1/2,+-1,+-i#.

Enjoy Maths.!

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