Question

Factorise the following expressions: 2x^2-x+1/8


Plz explain step by step
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Answer 1

Answer:

2x^2-x+1/8

=1/8(16x^2-8x+1)

=1/8{(4x)^2-2.4x.1+(1)^2}

=1/8(4x-1)^2

Answer 2

$❥\tt{ \: Solution:-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize ❥\footnotesize\tt\blue{ \: Factorise:-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\dashrightarrow \footnotesize \tt{ {2x}^{2} - x + \frac{1}{8}} \blue{ = 0 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize❥\footnotesize\tt \blue{ \: Taking \: LCM, \: LCM \: is \: 8 \: so \: multiply \: by \: 8 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\dashrightarrow \footnotesize \tt{ {(2 \times 8)x}^{2} - 8 \times x + \frac{1}{8} \times\frac{8}{1} } \blue{ = 0}$

$\dashrightarrow \footnotesize \tt{ {16x}^{2} - 8x + \frac{1}{ \cancel8} \times\frac{ \cancel8}{1} }\blue{ = 0}$

$\dashrightarrow \footnotesize \tt{ {16x}^{2} - 8x + 1 }\blue{ = 0}$

$\dashrightarrow \footnotesize \tt{ {16x}^{2} - 4x - 4x+ 1 }\blue{ = 0}$

$\dashrightarrow \footnotesize \tt{ 4 {x}^{2}(4x - 1 ) - 1(4x - 1) }\blue{ = 0}$

$\dashrightarrow \footnotesize \tt{(4x - 1 )(4x - 1) }\blue{ = 0}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\dashrightarrow \footnotesize \tt{(4x - 1 ) }\blue{ = 0} \: \: \: \: | \: \: \: \: (4x - 1) = \blue{0}$

$\footnotesize \boxed {\boxed{ \dashrightarrow \footnotesize \tt{x }\blue{ = \frac{1}{4} }}} \: \: \: \: \: \: \: \: \: \: \: \: | \: \: \: \: \: \: \: \: \: \: \: \footnotesize\boxed { \boxed{ \dashrightarrow \footnotesize \tt{ x= \blue{ \frac{1}{4} }}}}$