Originally Posted by

**cloud5** 5. Factorize p(x)=

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

= ±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

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

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

Deduce both the real roots of the equation

Do this the same way as I did above.

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