Question

Find the zeroes of the given cubic polynomial:
$\bf{x^3-6x^2-2x+12=0}$

Chapter Name ⇒ Polynomials

Class ⇒ 10th

Remember ⇒ No spamming!

Answer 1

$\large \pmb {\sf {\red {\bigstar ~~Question:-}}}$

Find the zeroes of the given cubic polynomial :-

$\sf { x^3-6x^2-2x+12=0}$

$\large \pmb {\sf {\red {\bigstar ~~Solution:-}}}$

We know that a cubic polynomial has three zeroes.

• Now, the factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12.

Here,

• $\sf p(x)=x^3-6x^2-2x+12=0$
• We have to find the zeroes by substituting the factors of 12 in x.

$\green {\longmapsto \pmb {\sf x=1}}$

$\sf {(1)^3 - 6( {1)}^{2} } - 2(1) + 12$

$\sf 1-6 - 2 + 12$

$\sf 5 ≠ 0$

Therefore, x=1 is not a zero.

$\green {\longmapsto\pmb {\sf x=-1} }$

$\sf({ - 1)}^{3} - 6( { - 1)}^{2} - 2( - 1) + 12$

$\sf - 1 - 6 + 2 + 12$

$\sf 7≠0$

Therefore, x=-1 is not a zero.

$\green {\longmapsto \pmb {\sf x=2}}$

$\sf {(2)}^{3} - 6( {2)}^{2} - 2( 2) + 12$

$\sf 8 - 24 - 4 + 12$

$\sf -8≠0$

Therefore, x=2 is not a zero.

$\green {\longmapsto \pmb{\sf x=-2 }}$

$\sf (-2)^3-6(-2)^2 - 2( - 2) + 12$

$\sf - 8 - 24 + 4 + 12$

$\sf - 16≠0$

Therefore, x=-2 is not a zero.

$\green {\longmapsto\pmb{\sf x=3}}$

$\sf(3)^3-6(3)^2-2(3)+12$

$\sf 27 - 54 - 6 + 12$

$\sf -21≠0$

$\green {\longmapsto \pmb {\sf x=-3}}$

$\sf (-3)^3-6(-3)^2-2(-3)+12$

$\sf - 27 - 54 + 6 + 12$

$\sf -63≠0$

Therefore, x=-3 is not a zero.

$\green {\longmapsto \pmb {\sf x=4 }}$

$\sf (4)^3-6(4)^2-2(4)+12$

$\sf 64-96 - 8 + 12$

$\sf -28≠0$

Therefore, x=4 is not a zero.

$\green {\longmapsto \pmb{\sf x=-4 }}$

$\sf (-4)^3-6(-4)^2-2(-4)+12$

$\sf - 64 - 96 + 8 + 12$

$\sf - 140≠0$

$\green {\longmapsto \pmb{\sf x=6 }}$

$\sf (6)^3-6(6)^2-2(6)+12$

$\sf 216-216 - 12 + 12$

$\sf 0≠0$

Therefore, x=6 is a zero.

Now, dividing the polynomial by x-6.

• Refer the attachment.

So, $\sf x^2-2$ is a factor of the polynomial.

$\sf x^2-2$

$\sf x=\pm \sqrt{2}$

$\large \pmb {\sf {\red {\bigstar Verification:-}}}$

Substituting x=6, ±√2 in the polynomial and solving :-

• Refer the attachment.

_____________________

Hence, x=6, +√2, -√2 are the zeroes of the polynomial.

Hope the concept's clear!

Answer 2

Answer:

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -2x+12

Group 2: -6x2+x3

Pull out from each group separately :

Group 1: (x-6) • (-2)

Group 2: (x-6) • (x2)

-------------------

Add up the two groups :

(x-6) • (x2-2)

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Factoring: x2-2

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 2 is not a square !!

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A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well

______________________

Solve : x2-2 = 0

Add 2 to both sides of the equation :

x2 = 2

When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:

x = ± √ 2

The equation has two real solutions

_________________

i Just Explained you the Method and the other hope it's understandable.