I expanded it, and by hit and trial method, -1 was the solution. Then I divided the expanded polynomial by (x+1). The resultant cubic can be easily factorised.

@Mục Xiên
–
well, we want to start to factorize the polynomial. You can use the rational root theorem
to guess at possible roots, and after which use the remainder factor theorem to help factorize. Have you seen these theorems before?

@Calvin Lin
–
Well, I think I haven't seen these theorems before. Because I'm a middle school students.
BTW, I'll ask my teacher for this problem's explaination, thanks for your helps!

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## Comments

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TopNewestI expanded it, and by hit and trial method, -1 was the solution. Then I divided the expanded polynomial by (x+1). The resultant cubic can be easily factorised.

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\( x = y + 1\)

\( ( y + 1)^2 + (\dfrac{y + 1}{y})^2 = \dfrac{5}{4}\)

\( y^2 + 2y + 1 + 1 + \dfrac{1}{y^2} + \dfrac{2}{y} = \dfrac{5}{4}\)

\( y^2 + \dfrac{1}{y^2} + 2y + \dfrac{2}{y} + 2 = \dfrac{5}{4}\)

\( a = y + \dfrac{1}{y}\)

\( a^2 - 2 + 2a + 2 = \dfrac{5}{4}\)

\( (a + 1)^2 = (\pm \dfrac{3}{2})^2\)

\( a = - \dfrac{5}{2} , \dfrac{1}{2}\)

\( y + \dfrac{1}{y} = - \dfrac{5}{2} , y + \dfrac{1}{y} = \dfrac{1}{2}\)

\( y = \dfrac{ -\dfrac{5}{2} \pm \sqrt{\dfrac{9}{4}}}{2} , y = \dfrac{ \dfrac{1}{2} \pm i\sqrt{\dfrac{15}{4}}}{2}\)

\( x - 1 = \dfrac{ -\dfrac{5}{2} \pm \sqrt{\dfrac{9}{4}}}{2} , x - 1 = \dfrac{ \dfrac{1}{2} \pm i\sqrt{\dfrac{15}{4}}}{2}\)

\( (x - 1= \dfrac{-1}{2} , -4) ~or~ x - 1 = \dfrac{ \dfrac{1}{2} \pm i\sqrt{\dfrac{15}{4}}}{2}\)

\( x = \dfrac{1}{2} , -3 , \dfrac{3 \pm i\sqrt{15}}{4}\)

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x= -1 or 1/2

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Can I know how to solve this, please?

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What have you tried?

Do you know how to clear denominators and factorize?

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\(4x^2\) - \(8x^3\) + \(3x^2\) + \(10x\) - \(5\) = \(0\)

What to do next?

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rational root theorem

well, we want to start to factorize the polynomial. You can use theto guess at possible roots, and after which use the remainder factor theorem to help factorize. Have you seen these theorems before?

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