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).?

Question

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

Caleb Alexander · Accepted Answer

Please see the explanation below.

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]}

Ariana Alvarez · Answer

# 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#.

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.!

Chloe Alexander · Answer

undefined To show that $x = \frac{1}{4}$ is one of the roots of the equation $4x^3 - x^2 - 4x + 1 = 0$, we substitute $x = \frac{1}{4}$ into the equation:

$$4\left(\frac{1}{4}\right)^3 - \left(\frac{1}{4}\right)^2 - 4\left(\frac{1}{4}\right) + 1$$

This simplifies to:

$$1 - \frac{1}{16} - 1 + 1 = 0$$

Therefore, when $x = \frac{1}{4}$, the equation equals zero, confirming that $x = \frac{1}{4}$ is one of the roots.

Now, to factorize $4x^3 - x^2 - 4x + 1$ completely, we can use the fact that $x = \frac{1}{4}$ is a root. We'll perform polynomial division or synthetic division by dividing $4x^3 - x^2 - 4x + 1$ by $x - \frac{1}{4}$. The quotient will give us a quadratic expression. Dividing the given cubic polynomial by $x - \frac{1}{4}$ should yield:

$$4x^3 - x^2 - 4x + 1 = (x - \frac{1}{4})(4x^2 + 1)$$

Now, we have $4x^2 + 1$. This quadratic expression doesn't factorize further over the real numbers, so we cannot factorize the given polynomial completely using real numbers.

To solve the equation $4x^3 - x^2 - 4x + 1 = 0$, we need to find all its roots. We have one root already ($x = \frac{1}{4}$). The other roots might be complex. We can use numerical methods or software to find the approximate values of the other roots.

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

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