Question

factorise
${2x}^{2} + 11 \sqrt{2} + 24$

Answer 1

$<b >Hey here is your answer$

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= ${2x}^{2} + 11 \sqrt{2}x + 24$

= $2{x}^{2} + 11 \sqrt{2}x + 24$

= $2{x}^{2} + 8 \sqrt{2}x + 3 \sqrt{2}x + 24$

= $2x ( x + 4 \sqrt{2} ) + 3 \sqrt{2} ( x + 4 \sqrt {2} )$

= $( x + 4 \sqrt {2} ) ( 2x + 3 \sqrt{2} )$

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hope this helps you

### BE BRAINLY ###

Answer 2

2x^2 + 11√2 + 24

= 2x^2 + 0x + ( 11√2 + 24 ) = 0

first find roots of above quadratic equation by using formula given below and then convert the obtained roots into factors.

x = - b +- √ ( b^2 - 4ac ) / 2a

here , a = 2 , b = 0 , c = 11√2 + 24

x = -( 0) +- √{ (0)^2 - 88√2- 192 } / 4

x = +- √ - 8 ( 11√2 + 24 ) / 4

x = +- 2 √2 √( 11√2 + 24 ) i / 4

4x = +- 2√2 ( 11√2 + 24 ) i

so Roots are

( 4x + 2 √2 ( 11 √2 + 24 ) i , ( 4x - 2√2 ( 11 √2 + 24 ) i

hence , factors

2x^2 + 11 √2 + 24= { 4x + 2√2 ( 11√2 + 24 ) i } { 4x - 2√2 ( 11 √2 + 24 ) i }


note :- √-1 = i (iota )