Factorize
6x⁴-5x³-38x²-5x+6
Please describe your best possible bi - quad reduction
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Answered by
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Hey there! Thanks for the awesome question.Looks complicated,but it is easy.
6x⁴-5x³-38x²-5x+6
=6x⁴+12x³-17x³-34x²-4x²-8x+3x+6
=(6x⁴+12x³)-(17x³+34x²)-(4x²+8x)+(3x+6)
=6x³(x+2)-17x²(x+2)-4x(x+2)+3(x+2)
=(x+2)(6x³-17x²-4x+3)
=(x+2)(6x³-18x²+x²-3x-x+3)
=(x+2)[(6x³-18x²)+(x²-3x)-(x-3)]
=(x+2)[(6x²(x-3)+x(x-3)-(x-3)]
=(x+2)(x-3)(6x²+x-1)
=(x+ 2)(x-3)(3x-1)(2x+1)
Hope helped! ^^
6x⁴-5x³-38x²-5x+6
=6x⁴+12x³-17x³-34x²-4x²-8x+3x+6
=(6x⁴+12x³)-(17x³+34x²)-(4x²+8x)+(3x+6)
=6x³(x+2)-17x²(x+2)-4x(x+2)+3(x+2)
=(x+2)(6x³-17x²-4x+3)
=(x+2)(6x³-18x²+x²-3x-x+3)
=(x+2)[(6x³-18x²)+(x²-3x)-(x-3)]
=(x+2)[(6x²(x-3)+x(x-3)-(x-3)]
=(x+2)(x-3)(6x²+x-1)
=(x+ 2)(x-3)(3x-1)(2x+1)
Hope helped! ^^
nethranithu:
Wow maths genius
hey bro. can you plz explain the approach to this question
we have just rearranged the terms to make sure we have like terms.
yup understood that much in the first go.. but how to determine the splitting of those times in one go
there is a small mistake in ur answer
see the last step (x+3) and last second step (x-3)
u had made mistake in this
Thanks supriya :)
ok
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