# Factorisation

Factorise : $$x^4 - 8x^2 + x + 12$$

Note by Dev Sharma
2 years, 10 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 $(x^2+x-4)(x^2-x-3)$

- 2 years, 6 months ago

Nice one...

- 2 years, 6 months ago

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

- 2 years, 6 months ago

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

# wolframalpha!

- 2 years, 10 months ago

with process?

- 2 years, 10 months ago