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Factorisation

Factorise : \(x^4 - 8x^2 + x + 12\)

Note by Dev Sharma
2 years, 2 months ago

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2 methods:\[\] 1)find the roots first. \[x^4-8x^2+16=4-x\] \[(x^2-4)^2=4-x\] \[x^2=4\pm\sqrt{4-x}\] \[x=\pm\sqrt{4+\sqrt{4-x}},\pm\sqrt{4-\sqrt{4-x}}\] second part first \[x=\sqrt{4-\sqrt{4-x}}=\sqrt{4-\sqrt{4-\sqrt{4-\sqrt{4-x}}}}=\sqrt{4-\sqrt{4-\sqrt{.....}}}\] \[x=\sqrt{4-x}\] \[x^2+x-4=0\] divide the polynomial by this by long division to find it is equal to \[(x^2+x-4)(x^2-x-3)\] 2)read this. we find a function of z whoch is cubic like in the note \[z^3-\frac{8}{2}z^2+\frac{8^2-4*12}{16}-\dfrac{1^2}{64}=z^3-4z^2+z-\dfrac{1}{64}\] by the rational root theorem we observe that \(\frac{1}{4}\) is a root. divide: \[(z-\dfrac{1}{4})(z^2-\dfrac{15}{4}z+\dfrac{1}{16})\] we can fid the sum of thee roots of second factor=\(\sqrt{z_1}+\sqrt{z_2}=\sqrt{z_1+z_2+2\sqrt{z_1z_2}}=\sqrt{\frac{15}{4}+2\sqrt{\dfrac{1}{16}}}=\dfrac{\sqrt{17}}{2}\) the negative of total sum =\(-\sqrt{\dfrac{1}{4}}-\dfrac{\sqrt{17}}{2}=-\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}\) we know that chances are \(-\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}\) is also a root.implying \[(x-(-\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}))(x-(-\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}))=x^2+x-4=0\] so, we divide by long division to find the polynomial equals \[(x^2+x-4)(x^2-x-3)\]

Aareyan Manzoor - 1 year, 10 months ago

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Nice one...

Dev Sharma - 1 year, 10 months ago

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thanks, did you read the linked note, i used it in second method, hope you like it.

Aareyan Manzoor - 1 year, 10 months ago

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(x^2-x-3)(x^2+x-4).

wolframalpha!

Nelson Mandela - 2 years, 2 months ago

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with process?

Dev Sharma - 2 years, 2 months ago

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