Factorise-m 4- m 2-1A- m 2- m -1 m 2- m -1B- m 2- m -1 m 2- m -1C- m 2- m -1 m 2- m -1D- m 2- m -1 m 2- m -1

The correct option is A (m^2 + m + 1)(m^2 m + 1)Given, the expression is nbsp; m^4 + m^2 +1. The above expression can be written as, =m^4 + 2m^2 m^2 +1 = ...

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Standard VIII

Mathematics

Factorisation Using Algebraic Identities

Factorise:m 4...

Question

Factorise:
$m^{4} + m^{2} + 1.$

(m^{2} + m + 1) (m^{2} - m + 1)

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(m^{2} - m + 1) (m^{2} - m + 1)

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(m^{2} - m - 1) (m^{2} - m + 1)

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(m^{2} - m - 1) (m^{2} - m - 1)

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Solution

The correct option is A $(m^{2} + m + 1) (m^{2} - m + 1)$
Given, the expression is
$m^{4} + m^{2} + 1.$
The above expression can be written as,
$= m^{4} + 2 m^{2} - m^{2} + 1$
$= m^{4} + 2 m^{2} + 1 - m^{2}$
$= (m^{2})^{2} + 2 m^{2} + 1^{2} - m^{2}$
Now, applying the identity
$(a + b)^{2} = a^{2} + 2 a b + b^{2},$ we get
$(m^{2} + 1)^{2} - m^{2}$
$[Using the identity: a^{2} - b^{2} = (a + b) (a - b)]$
$= (m^{2} + 1 + m) (m^{2} + 1 - m)$
$= (m^{2} + m + 1) (m^{2} - m + 1)$

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Similar questions

Q. If sec θ tan θ = m, show that

(\frac{m^{2} - 1}{m^{2} + 1}) = sinθ .

Q. If sec θ + tan θ = m, prove that sin

θ = \frac{(m^{2} - 1)}{(m^{2} + 1)} .

Q. If (cosec A + cot A) = m then prove that

\frac{m^{2} - 1}{m^{2} + 1} = \cos θ

Question 10

Write the correct answer from the given four options.

If one member of a Pythagorean triplet is $2 m$ , then the other two members are

a) $m, m^{2} + 1$
b) $m^{2} + 1, m^{2} - 1$
c) $m^{2}, m^{2} - 1$
d) $m^{2}, m + 1$

Q. Simplify the following expression:

(\frac{m + \sqrt{m^{2} - n^{2}}}{m - \sqrt{m^{2} - n^{2}}} - \frac{m - \sqrt{m^{2} - n^{2}}}{m + \sqrt{m^{2} - n^{2}}}) \frac{n^{2}}{4 m \sqrt{m^{2} - n^{2}}}

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Factorise: m4+m2+1.

Factorise:
$m^{4} + m^{2} + 1.$