Question

Obtain all the zeroes of the polynomial x⁴-7x³+17x²-17x+6, if two of it's zeroes are 3 and 1.

Answer 1

Answer:2 and 1

Step-by-step explanation: suppose the other zeros are a and b. The the equation will be :

(X-3)*(x-1)*(x-a)*(x-b) = x⁴-7x³+17x²-17x+6

(X^2-4x+3)*(x^2-(a+b)x +a*b)= x⁴-7x³+17x²-17x+6

X^4+x^3(-4-a-b)+x^2*(a*b+4a+4b+3)+x*(-4*a*b-3*a-3*b)+3*a*b= x⁴-7x³+17x²-17x+6

Which implies :

3*a*b = 6 or a*b= 2

-4*a*b -3*a-3*b =-17 or -4*2 -3*(a+b) =-17 or a+b = 3

A*b +4*(a+b)+3=17 or 2+4*(a+b)+3=17 or a+b = 3

Now we have 2 equations :

A+b = 3 and ab = 2

Solving this we get a=2, b=1

Answer 2

All the zeroes are 1, 1, 2, 3.

Step-by-step explanation:

The given polynomial has zeroes 3 and 1.

Thus (x - 3) and (x - 1) are its factors

i.e., (x - 3) (x - 1) is a factor

i.e., (x² - 4x + 3) is a factor

Now, x⁴ - 7x³ + 17x² - 17x + 6

= x⁴ - 4x³ + 3x² - 3x³ + 12x² - 9x + 2x² - 8x + 6

= x² (x² - 4x + 3) - 3x (x² - 4x + 3) + 2 (x² - 4x + 3)

= (x² - 4x + 3) (x² - 3x + 2)

So another factor is (x² - 3x + 2)

Now, x² - 3x + 2

= x² - x - 2x + 2

= x (x - 1) - 2 (x - 1)

= (x - 1) (x - 2)

Hence the other two zeroes are 1 and 2

∴ all the zeroes are 1, 1, 2, 3.

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