Algebra Theory - Common roots for polynomial equations

➔ Common root for polynomial equation

HCF / GCD (600, 136)

➔ Euclidean algorithm:

136) 600(4)‾−544‾15656) 136(2)‾−112‾12424) 56(2)‾−48‾188) 24(3)‾−24‾10\begin{align*} 136 & \overline{\smash{\big)}\, 600 \quad (4)} \\ & \underline{-544} \\ & \phantom{1}56 \\ \\ 56 & \overline{\smash{\big)}\, 136 \quad (2)} \\ & \underline{-112} \\ & \phantom{1}24 \\ \\ 24 & \overline{\smash{\big)}\, 56 \quad (2)} \\ & \underline{-48} \\ & \phantom{1}8 \\ \\ 8 & \overline{\smash{\big)}\, 24 \quad (3)} \\ & \underline{-24} \\ & \phantom{1}0 \\ \end{align*}

So, HCF = 8

➔ HCF (200, 150)

150) 200(1)‾−150‾15050) 150(3)‾−150‾10\begin{align*} 150 & \overline{\smash{\big)}\, 200 \quad (1)} \\ & \underline{-150} \\ & \phantom{1}50 \\ \\ 50 & \overline{\smash{\big)}\, 150 \quad (3)} \\ & \underline{-150} \\ & \phantom{1}0 \\ \end{align*}

So, HCF = 50

Example 1:

 Find common roots of

x3+x2−2x−2andx3−x2−2x+2x^3 + x^2 - 2x - 2 \quad \text{and} \quad x^3 - x^2 - 2x + 2

Solution using Euclid GCD method:

Polynomial Division:

x3−x2−2x+2x3+x2−2x−2⇒First term: 1\frac{x^3 - x^2 - 2x + 2}{x^3 + x^2 - 2x - 2} \Rightarrow \text{First term: } 1 x3−x2−2x+2−(x3+x2−2x−2)=−2x2+4x^3 - x^2 - 2x + 2 - \left(x^3 + x^2 - 2x - 2\right) = -2x^2 + 4 −2x2+4x3+x2−2x−2⇒Next term: −2x⋅(x3+x2−2x−2)=x3−2x\frac{-2x^2 + 4}{x^3 + x^2 - 2x - 2} \Rightarrow \text{Next term: } -\frac{2}{x} \cdot (x^3 + x^2 - 2x - 2) = x^3 - 2x

Now:

−2x2+4−(x3−2x)=x2−2-2x^2 + 4 - (x^3 - 2x) = x^2 - 2

Now divide:

−2x2−2⇒Subtraction gives remainder 0\frac{-2}{x^2 - 2} \Rightarrow \text{Subtraction gives remainder 0}

So,

x2−2 is the HCF⇒x2−2 is a common root factor⇒x=±2 are common rootsx^2 - 2 \text{ is the HCF} \Rightarrow x^2 - 2 \text{ is a common root factor} \Rightarrow x = \pm \sqrt{2} \text{ are common roots}

Example 2:

 Find common roots for

x4+5x3−22x2−50x+132=0(1)x^4 + 5x^3 - 22x^2 - 50x + 132 = 0 \tag{1} x4+x3−20x2+16x+24=0(2)x^4 + x^3 - 20x^2 + 16x + 24 = 0 \tag{2}

Polynomial Division Process

Divide (1) by (2):

x4+5x3−22x2−50x+132x4+x3−20x2+16x+24=1\frac{x^4 + 5x^3 - 22x^2 - 50x + 132}{x^4 + x^3 - 20x^2 + 16x + 24} = 1

Subtract:

(x4+5x3−22x2−50x+132)−(x4+x3−20x2+16x+24)=4x3+4x2−66x+108(x^4 + 5x^3 - 22x^2 - 50x + 132) - (x^4 + x^3 - 20x^2 + 16x + 24) = 4x^3 + 4x^2 - 66x + 108

Next division:

4x3−2x2−66x+108x3+x2−20x+24⇒Quotient term: 4xx\frac{4x^3 - 2x^2 - 66x + 108}{x^3 + x^2 - 20x + 24} \Rightarrow \text{Quotient term: } \frac{4x}{x}

Multiply and subtract:

4x(x3+x2−20x+24)=4x4+4x3−80x2+96x4x(x^3 + x^2 - 20x + 24) = 4x^4 + 4x^3 - 80x^2 + 96x

Subtracting gives:

−2x2+14x+12-2x^2 + 14x + 12

Continue dividing until:

HCF=4x2−20x+24=4(x2−5x+6)⇒Common roots: x=2,3HCF = 4x^2 - 20x + 24 = 4(x^2 - 5x + 6) \Rightarrow \text{Common roots: } x = 2, 3

Example 3:

Let:

X={−5,−4,−3,−2,−1,0,1,2,3,4,5}X = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}

Question:
Number of pairs (a,b)(a, b) such that the polynomials

x2+ax+bandx3+bx+ax^2 + ax + b \quad \text{and} \quad x^2 + bx + a

have a common real root, where a,b∈Xa, b \in X

Solution:
Divide the cubic polynomial with the quadratic one to get the remainder:
a - a.x^2
For this remainder to be 0(which implies common factor), a = 0 and b = anything.
Case 1: a = 0, b = anything. 

Following the Euclid method, again divide the divisor by this remainder to get the new remainder:
ax + b + 1
For this to be 0, a = 0, b = -1
Case 2: a = 0, b = -1

Next Remainder: (b + 1).x + a
For this to be zero,
Case 3: b = -1, a = 0 (Same as Case 2)

Next remainder: (b+1) - a^2/(b+1)
For this to be zero,
(b+1)^2 - a^2 = 0
=> (b+1-a)(b+1+a) = 0
=> Case 4,5
Case 4: b + a = -1
=> (a,b) = (-5,4)...(0,-1)(-1,0)...(4,-5) => Total 10 pairs but (0,-1) is also present.
Case 5: a - b = 1
=> (a,b) = (5,4)..(0,-1)..(-4,-5) => Total 10 pairs but (0,-1) is also present.
Enumerate Case 1:
a = 0, b = anything => 11 pairs including (0,-1).
But for the roots to be real, b has to be <= 0.
Otherwise x^2 + ax + b = 0 will have imaginary roots.
So we have to discount cases where b >0 => Total 6 cases remaining.
And this includes (0,-1).

So total: 10 + 10 + 6 = 26.
But we have to remove 2 since (0,-1) has been counted thrice.
So 26 - 2 = 24 is the correct answer.



 

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