Question

Find the zeros of polynomial x2+3x+2 and verify the relationship between zeros and cofficient​

Answer 1

Given :–

• A polynomial, p(x) = x² + 3x + 2.

To Find :–

• The zeroes of the given polynomial.
• Also verify the relationship between both the zeroes and the coefficient.

Solution :–

Given,

$\rightarrow$ p(x) = x² + 3x + 2

And the zeroes of the polynomial is the value of x where p(x) = 0.

Now,

• Putting p(x) = 0.

So,

$\hookrightarrow$ x² + 3x + 2 = 0

By splitting the middle term, we get

$\hookrightarrow$ x² + 2x + 1x + 2 = 0

$\hookrightarrow$ x(x + 2) + 1(x + 2) = 0

$\hookrightarrow$ (x + 2) (x + 1) = 0

$\hookrightarrow$ x + 2 = 0 ; x + 1 = 0

$\hookrightarrow$ x = –2 ; x = –1

So,

The value of x = –2 and –1

Therefore,

$\alpha$ = –2 and $\beta$ = –1 are the zeroes of the polynomial.

$\rightarrow$ p(x) = x² + 3x + 2

Comparing with ax² + bc + c.

Here,

• a = 1.
• b = 3.
• c = 2.

Verification :–

● Sum of zeroes = $\sf \dfrac{Coefficient \: of \: x}{Coefficient \: of \: {x}^{2}}$

I.e.,

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

• $\alpha$ = –2
• $\beta$ = –1
• b = 3
• a = 1

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

$\hookrightarrow - 2 - 1 = - 3$

$\hookrightarrow - 3 = - 3$

Verified

● Product of zeroes = $\dfrac{Constant \: term}{Coefficient \: of \: {x}^{2}}$

I.e.,

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

• $\alpha$ = –2
• $\beta$ = –1
• c = 2
• a = 1

$\hookrightarrow ( - 2) \times ( - 1) = \dfrac{2}{1}$

$\hookrightarrow 2 = 2$

Verified

Since,

The L.H.S. and the R.H.S. are equal.

So, the rationship between both the zeroes and the coefficient is verified.