Question

Write the integral zeroes of the following polynomials: i) (x – 3)(x – 7) ii) ( x + 1)(3x + 2)

Answer 1

I.(x-3)(x-7)

x²-7x-3x+21

x²-10x+21

ii.(X+1)(3x+2)

3x²+2x+3x+2

3x²+5x+2

Hope it will help you

Answer 2

Answer:

(i) The integer zeros of the polynomial  (x – 3)(x – 7) =  3 and 7

(ii)The integer zeros of the polynomial ( x + 1)(3x + 2))  = -1

Step-by-step explanation:

Given polynomials are

i) (x – 3)(x – 7)

ii) ( x + 1)(3x + 2)

To find,

The integer zeros of the given polynomials

Recall the concepts

Zeros of a polynomial p(x) are the values of the variable 'x' such that p(x) = 0

Solution

(i) Let p(x) = (x – 3)(x – 7)

p(x) = 0 ⇒(x – 3)(x – 7) = 0

⇒(x – 3)= 0 or  (x –7) = 0

⇒ x = 3, x = 7

p(x) = 0 ⇒ x = 3, x = 7

Hence the zeros of the polynomial (x – 3)(x – 7) are x = 3 and 7

Since both 3 and 7 are integers, we have the integer zeroes of the polynomial = 3,7

ii)Let q(x) =  ( x + 1)(3x + 2)

p(x) = 0 ⇒  ( x + 1)(3x + 2) = 0

⇒  ( x + 1)(3x + 2) = 0

⇒  ( x + 1) or (3x + 2) = 0

⇒  x = -1 or 3x = -2

⇒  x = -1 or x = $\frac{-2}{3}$

Hence the zeros of the polynomial ( x + 1)(3x + 2) are x = -1, $\frac{-2}{3}$

Since x =   $\frac{-2}{3}$, is not an integer, we have the integer zero of the polynomial is -1

Hence the required answer is

(i) The integer zeros of the polynomial  (x – 3)(x – 7) =  3 and 7

(ii)The integer zeros of the polynomial ( x + 1)(3x + 2))  = -1

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