Question

How many zereos does the polynomial (x -3)2 _4 can have, also fined.​

Answer 1

$\bold \green{\underline {Step\:by\: Step\: explanation}}:-$

$\huge\underline\bold\red{Given :}$

$⇒\mathrm{(x - 3)^2 - 4}$

$⇒\mathrm{x^2 + 9 - 6x -4}$

$⇒\mathrm{x^2 - 6x + 5}$

As the equation has degree 2, then the zereos of the polynomial will be 2.

$⇒\mathrm{x^2 - 6x + 5 = 0}$

$⇒\mathrm{x^2 - x - 5x + 5 = 0}$

$⇒\mathrm{x(x -1) -5(x +1)}$

$⇒ \mathrm{(x - 1)} \mathrm{(x - 5)}$

$⇒ <strong>\</strong><strong>b</strong><strong>o</strong><strong>x</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>x = 1</strong><strong>}</strong><strong>\: </strong><strong>{</strong><strong>or</strong><strong>}</strong><strong>\: </strong><strong>\</strong><strong>b</strong><strong>o</strong><strong>x</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>x = 5</strong><strong>}</strong>$

Answer 2

$\huge\underline\mathrm{Correct\:question-}$

How many zeroes does the polynomial (x-3)² - 4 have? Also find its zeroes.

$\huge\underline\mathrm{Solution-}$

We know that,

★ (a-b)² = a² + b² - 2ab

$\mapsto$ [x² + (3)² - 2(x)(3)] - 4

$\mapsto$ [x² + 9 - 6x] - 4

$\mapsto$ x² + 9 - 6x - 4

$\mapsto$ x² - 6x + 5

Equate it to zero.

$\mapsto$ x² - 6x + 5 = 0

By splitting middle term,

$\mapsto$ x² - 5x - x + 5 = 0

$\mapsto$ x(x - 5) - 1(x - 5) = 0

$\mapsto$ (x - 1)(x - 5) = 0

$\mapsto$ x = 1 or 5

$\therefore$ Given Polynomial has two zeroes that are 1 and 5.