Question

Find the zeroes of the given polynomials.

(i) p(x) = (x + 2) (x + 3)
(ii) p(x) = x⁴ - 16​

Answer 1

for zeroes of the polynomial,

p(x)=0,

x⁴-16=0,

(x²)²-(2²)²=0,

(x²-2²)(x²+2²)=0,

(x²-4)(x²+4)=0,

if

x²-4=0,

then

x²-4=0,

x²=4,

x=√4,

x=+-2,

if

x²+4=0,

then

x²=-4,

x=√(-4),

x=2i

Answer 2

✰ Qᴜēsᴛíõñ :-

Find the zeroes of the given polynomials.

(i) p(x) = (x + 2) (x + 3)

(ii) p(x) = x⁴ - 16

✪ Söʟúᴛîøɴ :-

(i) Given p(x) = (x + 2) (x + 3)

It is a quadratic polynomial

It has almost two zeroes

Let p(x) = 0

➙ (x + 2) (x + 3) = 0

➙ (x + 2) = 0 or (x + 3) = 0

➙ x = -2 or x = -3

⛬ The zeroes of the polynomial are -2 and -3.

(ii) Given p(x) = x⁴ - 16 is a biquadratic polynomial. It has atmost two zeroes

Let p(x) = 0

➙ x⁴ - 16 = 0

➙ (x²)² - 4² = 0

➙ (x² - 4) (x² + 4) = 0

➙ (x + 2) (x - 2) (x² + 4) = 0

➙ (x + 2) = 0 or (x - 2) = 0 or (x² + 4) = 0

➙ x = -2 (or) x = 2 (or) x² = -4

x = + √-4

⛬ The zeroes of the polynomial are 2, -2, +√-4

We don't consider √-4. Since it is not real.