Question

Write the zeroes of the polynomial $\sf x^{2} - x - 6$

Answer 1

x2-x-6

x2-3x+2x-6

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

x = 3 or X=-2

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Answer 2

Answer:

$\rightarrow\small\bf x = -2 \: and \: x = 3$

Step-by-step explanation:

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• Write the zeroes of the polynomial

$\: \: \: \: \: \: \: {x}^{2} - \: x \: - 6$

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• We will find the required zeroes of the polynomial p(x) by factorising it by using the method of the splitting the middle term.

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$p(x) = {x}^{2} \: - x \: - 6$

• Here; a = 1, b = -1 and c = -6

a × c = 1 × -6 = -6

• We need factors of -6 such that by adding or subtracting them we get 1 and by multiplying both of them we get -6.

we have;

• -6 = 1 × -6
• -6 = -2 × 3
• -6 = 2 × -3

Here we can see that -2 and 3 are the most suitable as -2 × 3 = -6 and -2+3 = -1.

Now,

Using the middle term splitting method we have,

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

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

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

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

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

Therefore, the factors of the polynomial are (x+2)(x-3)

$\mathcal{\green{And}}$

The zeroes are

$(x - 3) = 0 = > 3$

$\mathcal{\green{And}}$

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

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For p(x);

If p(a) = 0 then x = 0 is a zero of polynomial.

Let's verify by putting p(-2)

${ ( - 2)}^{2} - ( - 2) - 6$

$4 + 2 - 6$

$= 0$

$\mathcal{\green{And}}$

Put p(3)

${(3)}^{2} - (3) - 6$

$9 - 3 - 6$

$9 - 9 = 0$

Hence, verified.

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{More \ to \ know:-}}\mid}}}}$

• Zeros of a polynomial can be defined as the points where the polynomial becomes zero as a whole.

• A polynomial having value zero (0) is called zero polynomial.

• The degree of a polynomial is the highest power of the variable x.

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