Question

please answer question9​

Answer 1

Step-by-step explanation:

Given -

• √3 is a zero of the polynomial p(x) = x³ + x² - 3x - 3

To Find -

• Other zeroes

Now,

→ x³ + x² - 3x - 3

→ x³ - √3x² + (√3x² + x²) - (√3x + 3x) + √3x - 3

→ x²(x - √3) + (√3x + x)(x - √3) + √3(x - √3)

→ (x² + √3x + x + √3)(x - √3)

→ (x² + x + √3x + √3)(x - √3)

→ [x(x + 1) + √3(x + 1)](x - √3)

→ (x - √3)(x + 1)(x + √3)

Zeroes are -

→ x - √3 = 0 and x + 1 = 0 and x + √3 = 0

→ x = √3 and x = -1 and x = -√3

Answer 2

Given:-

• p(x) = x³ + x² -3x -3
• √3 is zeroes of the polynomial

To Find:-

• Other two zeroes

Solution:-

x³ + x² -3x -3

x³ - √3x² + (√3x² + x²) - (√3x + 3x) + √3x - 3

x²(x - √3) + (√3x + x)(x - √3) + √3(x - √3)

(x² + √3x + x + √3)(x - √3)

(x² + x + √3x + √3)(x - √3)

[x(x + 1) + √3(x + 1)][(x - √3)]

(x - √3)(x + 1)(x + √3)

★ Hence, zeroes of the polynomial are:-

• x - √3 = 0
• x = √3

★ and

• x + 1 = 0
• x = -1

★ and

• x + √3 = 0
• x = - √3

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