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Factorise-3x-4-6x-3-8x-2-2x-3-0-

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Question Number 46378 by rahul 19 last updated on 24/Oct/18
Factorise : 3x^4 +6x^3 +8x^2 −2x−3=0.
$${Factorise}\:: \\ $$$$\mathrm{3}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0}. \\ $$
Answered by MJS last updated on 24/Oct/18
factor 1 =3 x^4 +2x^3 +(8/3)x^2 −(2/3)x−1=0 (x^2 +ax+b)(x^2 +cx+d)=0 by comparing the constants we get 4 equations (1) a+c−2=0 (2) ac+b+d−(8/3)=0 (3) ad+bc+(2/3)=0 (4) bd+1=0 solving (1) for a, (2) for b, (3) for d we get (4) 9c^6 −54c^5 +156c^4 −364c^3 +280c^2 −176c=0 ⇒ c=0, a=2, b=3, d=−(1/3) so we have 3(x^2 +2x+3)(x^2 −(1/3))=0 solving the square factors we get 3(x+1−i(√2))(x+1+i(√2))(x−((√3)/3))(x+((√3)/3))=0
$$\mathrm{factor}\:\mathrm{1}\:=\mathrm{3} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\frac{\mathrm{8}}{\mathrm{3}}{x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{ax}+{b}\right)\left({x}^{\mathrm{2}} +{cx}+{d}\right)=\mathrm{0} \\ $$$$\mathrm{by}\:\mathrm{comparing}\:\mathrm{the}\:\mathrm{constants}\:\mathrm{we}\:\mathrm{get}\:\mathrm{4}\:\mathrm{equations} \\ $$$$\left(\mathrm{1}\right)\:\:{a}+{c}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:{ac}+{b}+{d}−\frac{\mathrm{8}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:{ad}+{bc}+\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:{bd}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solving}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{a},\:\left(\mathrm{2}\right)\:\mathrm{for}\:{b},\:\left(\mathrm{3}\right)\:\mathrm{for}\:{d}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{4}\right)\:\:\mathrm{9}{c}^{\mathrm{6}} −\mathrm{54}{c}^{\mathrm{5}} +\mathrm{156}{c}^{\mathrm{4}} −\mathrm{364}{c}^{\mathrm{3}} +\mathrm{280}{c}^{\mathrm{2}} −\mathrm{176}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{c}=\mathrm{0},\:{a}=\mathrm{2},\:{b}=\mathrm{3},\:{d}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\mathrm{solving}\:\mathrm{the}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{3}\left({x}+\mathrm{1}−\mathrm{i}\sqrt{\mathrm{2}}\right)\left({x}+\mathrm{1}+\mathrm{i}\sqrt{\mathrm{2}}\right)\left({x}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\left({x}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)=\mathrm{0} \\ $$

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