Question

Please tell the answer of this by factor theorem

Answer 1

To factorise -

Factorise : 27p³ - 1 / 216 - 9/2 p² + 1/4 p

Solution -

Method 1 { Factor Theorem } -

Given G ( p ) = 27p³ - 1 / 216 - 9/2 p² + 1/4 p

Substitute the value of p = { 1 / 18 }

Now find the value of G { 1 / 18 }

G { 1 / 18 } -

=> 27 × [ 1 / 18 ]³ - 1 / 216 - 9/2 × [ 1 / 18 ]² + [ 1 / 4 ] × [ 1 / 18 ]

=> 27 × [ 1 / 5832 ] - [ 1 / 216 ] - [ 9 / 2 ] × [ 1 / 324 ] + [ 1 / 72 ]

=> [ 27 / 5832 ] - [ 1 / 216 ] - [ 9 / 648 ] + [ 1 / 72 ]

=> 0

So , we can say that -

( a - 1 / 18 ) is a factor of this polynomial .

This means that -

Multiplying by 2 both sides,

( 3a - 1 / 6 ) is a factor of this polynomial .

However , solving by this method is unnecessarily complicated.

It is better and easier to factorise normally .

Method 2 { Normal Method } -

Given G ( p ) = 27p³ - 1 / 216 - 9/2 p² + 1/4 p

=> G ( p ) = [ 3 p ] ³ - 9/2 p² + 1/4 p - [ 1 / 6 ]³

This is clearly in the form of ( a - b )³

Final Answer -

Factorising , we get -

27p³ - 1 / 216 - 9/2 p² + 1/4 p = { 3a - 1/6 }³

_________________________________________________

Answer 2

Question :–

Factorise: $\sf {27p}^{3} - \dfrac{1}{216} - {\dfrac{9}{2} p}^{2} + \dfrac{1}{4}p$

Solution :–

→ $\sf {27p}^{3} - \dfrac{1}{216} - {\dfrac{9}{2} p}^{2} + \dfrac{1}{4}p$

→ $({3p})^{3} - { (\frac{1}{6} )}^{3} - 3 {(3p)}^{2} (\frac{1}{6} ) + 3(3p)( { \frac{1}{6} )}^{2}$

We know that,

→ $(x - y) = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3}$

Here,

• x = 3p

• y = $\sf \frac{1}{6}$

After comparing them the expressions becomes,

→ $( {3p)}^{3} - { (\frac{1}{6}) }^{3} - { (\frac{9}{2} )}^{2} + \frac{1}{4} p = {(3p - \frac{1}{6} )}^{3}$

In factored form :–

→ $(3p - \frac{1}{6} )(3p - \frac{1}{6} )(3p - \frac{1}{6} )$

Hence,

The factorisation of:

${27p}^{3} - \frac{1}{216} - {\frac{9}{2} p}^{2} + \frac{1}{4}p \:$ is $\: (3p - \frac{1}{6} )(3p - \frac{1}{6} )(3p - \frac{1}{6} )$