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**cloud5** 5. Factorize p(x)=$\displaystyle 8x^4-4x^3-2x^2+7x-3$ completely. Show that the equation has only two real roots, and find these both roots.

Now that's more like it...

5) By the rational roots theorem, the possible rational roots are

$\displaystyle \frac{\text{factors of}\; 3}{\text{factors of}\; 8} = \frac{\pm 1, \pm 3}{\pm 1, \pm 2, \pm 4, \pm 8}$

= ±1, ±3, ±1/2, ±3/2, ±1/4, ±3/4, ±1/8, ±3/8.

Start testing these factors by using the Factor Theorem or synthetic division. You'll find that x = -1 is a root:

Code:

-1| 8 -4 -2 7 -3
--- -8 12 -10 3
----------------------
8 -12 10 -3 0

x = 1/2 is also a root:

Code:

1/2| 8 -12 10 -3
---- 4 -4 3
-------------------
8 -8 6 0

So

$\displaystyle \begin{aligned}

p(x) &= 8x^4-4x^3-2x^2+7x-3 \\

&= (x - 1/2)(x + 1)(8x^2 - 8x + 6) \\

&= 2(x - 1/2)(x + 1)(4x^2 - 4x + 3) \\

&= (2x - 1)(x + 1)(4x^2 - 4x + 3)

\end{aligned}$

The quadratic is irreducible over the reals -- the discriminant is negative:

$\displaystyle b^2 - 4ac = (-4)^2 - 4(4)(3) = 16 - 48 = -32$

We have two real roots, x = 1/2 and x = -1.

Deduce both the real roots of the equation $\displaystyle x^4-x^3-x^2+7x-6=0$

Do this the same way as I did above.

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