Question

) If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find all the zeroes​

Answer 1

Answer:

We are given with the polynomial here,

• p(x) = x3-3x2+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴ px³ + qx² +rx+s = x³ - 3x² + x + 1

• p = 1

• q = -3

• r = 1

• s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

• Putting the values q and p.

-(-3)/1 = 3a

a=1

• Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

➞ -s/p = 1-b²

➞ -1/1 = 1-b²

➞ b2 = 1+1 = 2

➞ b = √2

Hence,1-√2, 1, 1+√2 are the zeroes of x3- 3x2 +x +1.

Answer 2

Answer:

We are given with the polynomial here,

• p(x) = x3-3x2+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴ px³ + qx² +rx+s = x³ - 3x² + x + 1

• p = 1

• q = -3

• r = 1

• s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

• Putting the values q and p.

-(-3)/1 = 3a

a=1

• Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

➞ -s/p = 1-b²

➞ -1/1 = 1-b²

➞ b2 = 1+1 = 2

➞ b = √2

Hence,1-√2, 1, 1+√2 are the zeroes of x3- 3x2 +x +1.

Answer 3

Answer:

We are given with the polynomial here,

• p(x) = x3-3x2+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴ px³ + qx² +rx+s = x³ - 3x² + x + 1

• p = 1

• q = -3

• r = 1

• s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

• Putting the values q and p.

-(-3)/1 = 3a

a=1

• Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

➞ -s/p = 1-b²

➞ -1/1 = 1-b²

➞ b2 = 1+1 = 2

➞ b = √2

Hence,1-√2, 1, 1+√2 are the zeroes of x3- 3x2 +x +1.