Question

The zeroes of the polynomial p(x) = x² +x-6 are
(a) 2,3
(b)-23
(c) 2.-3
(d) -2,-3​

Answer 1

$\huge \underline { \bf { \red S \green O \pink L \blue U \orange T \purple I \red O \pink N \green{...}}}$

By splitting middle term :

$:\implies \sf {x}^{2} + x - 6 = 0 \\ \\ :\implies \sf {x}^{2} + 3x - 2x - 6 = 0 \\ \\:\implies \sf x(x + 3) - 2(x + 3) = 0\\ \\:\implies \sf (x - 2)(x + 3) = 0 \\ \\:\implies \sf x - 2 = 0 \: \: , \: \: x + 3 = 0 \\ \\:\implies \boxed{\bf x = 2 }\: \: , \: \: \boxed{ \bf x = - 3}$

$\therefore \underline{ \bf \blue{2 , -3 \: are \: the \: zeroes \: of \: the \: polynomial}}$

Answer 2

Given:

• $\sf{Polynomial \: p(x) = {x}^{2} + x - 6}$

To Find:

• Zeroes of the polynomial = ?

Solution:

First of all let's convert the given polynomial equation into a quadratic equation form:

$\\ \longrightarrow\rm{ {x}^{2} + x - 6} \\ \\ \longrightarrow\rm{ {x }^{2} + x - 6 = 0}$

We can do find the zeroes of the polynomial within two methods:

• Splitting middle term method.

• Factorization method.

Let us do it according by splitting the middle term of the polynomial:

$\\ \longrightarrow\bf{ {x}^{2} + x - 6= 0} \\ \\ \longrightarrow\bf{ {x}^{2} + 3x - 2x - 6 = 0 \: .... \: (Splitting \: middle \: terms)} \\ \\ \longrightarrow\bf{x(x + 3) - 2(x + 3) = 0} \\ \\ \longrightarrow\bf{(x - 2)(x + 3) = 0 \: .... \: (Collecting \: like \: terms)} \\ \\ \longrightarrow\bf{x - 2 = 0 \: or \: x + 3 = 0} \\ \\ \longrightarrow\bf{x = 0 + 2 \: , \: x = 0 - 3} \\ \\ \longrightarrow\bf{x = 2 \: , \: x = - 3}$

Thus,

$\therefore$ The zeroes of polynomial are 2 & -3.