Question

find the zeros of the polynomial x square minus x minus 6 and verify the relationship between the zeros and coefficients of the polynomial​

Answer 1

Answer:

given in step by step explantion

Step-by-step explanation:

using method:

$splitting \: the \: middle \: term$

$x {}^{2} - x - 6 = 0$

$x {}^{2} + 2x - 3x - 6$

$x(x + 2) - 3(x + 2) = 0$

$(x + 2)(x - 3) = 0$

The two zeroes are

$x = - 2 \: and \: x = 3$

$\boxed{Verification}$

sum of the zeroes

$\alpha + \beta = - 2 + 3 = 1 = \frac{ - (coefficient \: of \: x {}^{2}) }{coefficient \: of \: x} = \frac{ - (1)}{ - 1} = 1$

product of the zeroes

$\alpha \beta = - 2 \times 3 = - 6 = \frac{constant \: term}{coefficient \: of \: x {}^{2} } = \frac{ - 6}{1} = - 6$

Hence

$\boxed{Verified}$

Answer 2

The zeros of the polynomial are -2 and 3.

Step-by-step explanation:

Consider the provided polynomial.

$x^2-x-6=0$

Simplify the polynomial as shown below:

$x^2-x-6=0\\x^2-3x+2x-6=0\\x(x-3)+2(x-3)=0\\x=-2\ or x=3$

For the quadratic polynomial  

ax²+bx+c=0  

the relation between the coefficient a, b and c and their roots α and β is.

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

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

Here α = -2 and β=3.

Now verify it by substituting the respective values.

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

Which is true.

$-2\times3=\frac{-6}{1}=-6$

Which is true.

The zeros of the polynomial are -2 and 3.

#Learn more

Find the zeroes of the polynomial x^2+1/6x-2 and verify the relation between the coefficient and zeroes of the polynomial.

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