Thread: Polynomials questions again... part 3

1. Polynomials questions again... part 3

5. Factorize p(x)= $8x^4-4x^3-2x^2+7x-3$ completely. Show that the equation has only two real roots, and find these both roots.
Deduce both the real roots of the equation $x^4-x^3-x^2+7x-6=0$

6. Using the substitution $u=x+\frac{1}{x}$, find the values of a and b such that $x^2-5x+6-\frac{5}{x}+\frac{1}{x^2}=u^2+au+b$. Subsequently, determine all the roots of the equation $x^4-5x^3+6x^2-5x+1=0$

5. p(x)=(2x-1)(x+1)( $4x^2-4x+3)$; x=-1, x= $\frac{1}{2}$

6. a=-5, b=4; x=2 $\pm \sqrt{3}$, $x=\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$

2. Originally Posted by cloud5

6. Using the substitution $u=x+\frac{1}{x}$, find the values of a and b such that $x^2-5x+6-\frac{5}{x}+\frac{1}{x^2}=u^2+au+b$.

$x^2-5x+6-\frac{5}{x}+\frac{1}{x^2}$
$=x^2+\frac{1}{x^2}-5x-\frac{5}{x}+6$
$=(x+\frac{1}{x})^2-2-5(x+\frac{1}{x})+6$
$=u^2-2-5u+6$
$=u^2-5u+4$

hence, a=-5, b=4

Subsequently, determine all the roots of the equation $x^4-5x^3+6x^2-5x+1=0$
$x^4-5x^3+6x^2-5x+1=0$

Let x is not equal to zero.
$x^2(x^2-5x^1+6-\frac{5}{x}+\frac{1}{x^2})=0$
$x^2(u^2-5u+4)=0$

Solving for u gives $u=1,4$
so, $x+\frac{1}{x}=1,4$

Solving by quadratic formula, you will get

x=2 $\pm \sqrt{3}$, $x=\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$

3. Originally Posted by cloud5
5. Factorize p(x)= $8x^4-4x^3+ax^2+bx-3$ completely. Show that the equation has only two real roots, and find these both roots.
[snip]
I don't see how you can expect this equation to be factorised when the coefficients of $x^2$ and $x$ are not known ....

4. Originally Posted by mr fantastic
I don't see how you can expect this equation to be factorised when the coefficients of $x^2$ and $x$ are not known ....

5. Originally Posted by cloud5
5. Factorize p(x)= $8x^4-4x^3-2x^2+7x-3$ 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
$\frac{\text{factors of}\; 3}{\text{factors of}\; 8} = \frac{\pm 1, \pm 3}{\pm 1, \pm 2, \pm 4, \pm 8}$
= ±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
\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:
$b^2 - 4ac = (-4)^2 - 4(4)(3) = 16 - 48 = -32$
We have two real roots, x = 1/2 and x = -1.

Deduce both the real roots of the equation $x^4-x^3-x^2+7x-6=0$
Do this the same way as I did above.

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