Question

The zeroes of polynomial p(x)=x²-2x-3 are

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A polynomial
• X²-2x-3

${\sf{\green{\underline{\large{To\:Verify}}}}}$

• . zeroes of polynomial

${\sf{\pink{\underline{\Large{Explanation}}}}}$

x²-2x-3

• Spilt the middle term in such a way that sum become -2 and product 3

=>x²-2x-3

=>x²-3x+x-3

=>x(x-3)+1(x-3)

=>(x-3) (x+1)

Additional Information

Let the zeroes of the polynomial be$\tt\alpha{and}\beta$

Then,

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

&

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

Here,

a=1

b-2

C=-3

☞

$\tt\alpha{+}\beta{=}\dfrac{{2}}{1}=\dfrac{-b}{a}$

$\tt\alpha{+}\beta{=}\dfrac{Cofficient\:of\:X}{Cofficient\:of\:x^2}=$

&

$\tt\alpha{\times}\beta{=}\dfrac{-3}{1}$

$\tt{\large\alpha{\times}\beta{=}\dfrac{Constant\:term}{Cofficient\:of\:x^2}}$

Hence,relation verified

Answer 2

✯✯ QUESTION ✯✯

The zeroes of polynomial p(x)=x² - 2x - 3 are?

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

$\longmapsto\tt{{x}^{2}-2x-3}$

$\longmapsto\tt{{x}^{2}-(3x-1x)-3}$

$\longmapsto\tt{{x}^{2}-3x+1x-3}$

$\longmapsto\tt{x(x-3)+1(x-3)}$

$\longmapsto\tt{(x+1)(x-3)}$

• x = -1
• x = 3

➙So , -1 and 3 are the zeroes of Polynomials x²-2x-3..

➥Here : -

• a = 1
• b = -2
• c = -3

★Sum of Zeroes : -

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

$\longmapsto\tt{-1+3=\dfrac{-(-2)}{1}}$

$\longmapsto\tt{2=2}$

$\pink\longmapsto\:\large\underline{\boxed{\bf\red{L.H.S}\green{=}\orange{R.H.S}}}$

_______________________

★Product of Zeroes : -

$\longmapsto\tt{\alpha\beta=\dfrac{c}{a}}$

$\longmapsto\tt{-1\times{3}=\dfrac{-(-3)}{1}}$

$\longmapsto\tt{-3=-3}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{L.H.S}\orange{=}\purple{R.H.S}}}$

✺ HENCE VERIFIED✺