Question

find the zeros of the polynomial p(x) = x2 - X - 6.

Answer 1

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p(x)=x^2-x-6

=x^2-3x+2x-6

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

=(x+2)(x-3)

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

x=-2,3
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Answer 2

Answer:

Zeros of the given polynomial p(x) are 3 and (-2).

Step-by-step explanation:

Given polynomial is $p(x) = {x}^{2} - x - 6$

If $p(x) = 0$

Then,

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

By middle term method,

${x}^{2} - (3 - 2)x - 6 = 0$

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

If product of two number is zero then they are separately also zero.

So,$(x - 3) = 0 \\ x = 3$and $(x + 2) = 0 \\ x = - 2$

So Zeros of given polynomial p(x) are 3 and (-2).

This is a problem of Algebra,

Some important formulas of Algebra,

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

${a}^{2} - {b}^{2} = (a + b)(a - b)\\{a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab\\{a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab\\{a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )\\{a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$