Question

Find the zeroesof the polynomial x^-6x-27 and verify ​

Answer 1

Answer:

Step-by-step explanation:

ANSWER:-

Correct Polynomial :-

$\boxed{\sf{x^2 -6x - 27 }}$

So, we will first Factorise the Polynomial .

We will use the middle term Factorisation method.

• Terms of 27 are
• 3 and 9
• -3 and -9
• -9 and 3
• -3 and 9

$\sf{ x^2 -9x+3x-27}$ is the suitable term.

So now moving forward by taking common terms,

$\sf{x(x-9)+3(x-9)}\\\sf{(x+3)(x-9) are \ the \ suitable \ terms}$

$\boxed{\bf{x = -3 \ and \ 9}}$

VERIFICATION:-

Now as we have got two different Equations, let us assume

$\sf{ \alpha = -3 \ and \ \beta = 9}$

We know that:-

$\alpha + \beta = \frac{-b}{a}$

$\alpha\beta = \frac{c}{a}$

• Where a is 1
• Where b is 6
• Where c is 27

So now , we will place the numbers and verify.

With addition:-

$\sf{-3+9 \ | \ -(-6)}$

$\sf{ -6 \ | -6 \longrightarrow Matched }$

With Multiplication:-

$\sf{-3 \times 9 \ | -27}$

$\sf{-27 \ | \ -27 \longrightarrow Matched}$

Hence Verified!

Answer 2

$\sf\huge\green{\underline{ Question : }}$

Find the zeroes of the polynomial

p(x) = x² - 6x - 27 and verify.

$\sf\huge\green{\underline{ Solution : }}$

★ Given that,

• Polynomial p(x) = x² - 6x - 27

★ To find,

• Zeroes of the polynomial with verification.

★ Now,

Factories the polynomial.

$\rm\:\implies x^{2} - 6x - 27 = 0$

$\rm\:\implies x^{2} - 9x + 3x - 27 = 0$

$\rm\:\implies x(x - 9) + 3(x - 9) = 0$

$\rm\:\implies (x - 9)(x + 3) = 0$

$\rm\:\implies x = 9 , - 3$

$\boxed{\blue{ Hence, \:the\:zeroes\:are\:= 9, - 3}}$

★ Let,

Compare the polynomial p(x) with ax² + bx + c = 0

• a = 1
• b = - 6
• c = - 27

★ Now,

$\rm\:\implies Sum\:of\:the\:zeroes\:= 9 +(-3) = 9 - 3 = 6$

$\rm\: \implies \alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{1} = 6$

$\rm\:\implies Product\:of\:the\:zeroes\:= 9(-3) = - 27$

$\rm\:\implies \alpha\beta = \frac{c}{a}= \frac{-27}{1}=-27$

$\underline{\boxed{\bf{\purple{ \therefore Hence,\:it\:is\:verified}}}}\:\orange{\bigstar}$