Question

Zeroes of the polynomial :

X^2 + 7X + 10

X^2 - 25

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Answer 1

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1)

$= > x {}^{2} + 7x + 10 = 0 \\ = > {x}^{2} + 2x + 5x + 10 = 0 \\ = > x(x + 2) + 5(x + 2) = 0 \\ = > (x + 2)(x + 5) = 0 \\ = > x + 2 = 0 \: \: \: \: or \: \: x + 5 = 0 \\ = > x = - 2 \: \: \: \: \: or \: \: x = - 5$

2)

$= > x {}^{2} - 25 = 0 \\ = > {x}^{2} = 25 \\ = > x = \sqrt{25} \\ = > x = 5 \: \: \: \: \: or \: \: \: x= - 5$

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Answer 2

Solution :-

i) x² + 7x + 10

First factorise it

x² + 7x + 10

Splitting the middle terms

Multiply first term and last terms

x² * 10 = 10x²

Find factors of 10x² which sums to middle term i.e 7x

• x * 10x

• 2x * 5x

So 2x and 5x are the factor which sums to middle term i.e 7x(2x + 5x = 7x)

Repalce 7x with 2x + 5x in the above expression

= x² + 2x + 5x + 10

= x(x + 2) + 5(x + 2)

= (x + 2)(x + 5)

Now equate the factors of x² + 7x + 10 i.e (x + 2)(x + 5) to 0 to get the zeroes

⇒ (x + 2)(x + 5) = 0

⇒ x + 2 = 0

x + 5 = 0

⇒ x = - 2

x = - 5

- 5, - 2 are the zeroes of polynomial x² + 7x + 10.

ii) x² - 25

First factorise it

= (x)² - 5²

= (x + 5)(x - 5)

[Because x² - y² = (x + y)(x - y)]

Now equate the factors of x² - 25 i.e (x + 5)(x - 5) to 0 to get the zeroes

⇒ (x + 5)(x - 5) = 0

⇒ x + 5 = 0

x - 5 = 0

⇒ x = - 5

x = 5

- 5, 5 are the zeroes of polynomial x² - 25.