Find the zeroes of the given cubic polynomial:
[tex]\bf{x^3-6x^2-2x+12=0}[/tex]
Chapter Name ⇒ Polynomials
Class ⇒ 10th
Remember ⇒ No spamming!
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Find the zeroes of the given cubic polynomial:
[tex]\bf{x^3-6x^2-2x+12=0}[/tex]
Chapter Name ⇒ Polynomials
Class ⇒ 10th
Remember ⇒ No spamming!
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[tex]\large \pmb {\sf {\red {\bigstar ~~Question:-}}}[/tex]
Find the zeroes of the given cubic polynomial :-
[tex]\sf { x^3-6x^2-2x+12=0}[/tex]
[tex] \\ [/tex]
[tex]\large \pmb {\sf {\red {\bigstar ~~Solution:-}}}[/tex]
We know that a cubic polynomial has three zeroes.
Here,
[tex]\green {\longmapsto \pmb {\sf x=1}}[/tex]
[tex]\sf {(1)^3 - 6( {1)}^{2} } - 2(1) + 12[/tex]
[tex]\sf 1-6 - 2 + 12[/tex]
[tex]\sf 5 ≠ 0[/tex]
Therefore, x=1 is not a zero.
[tex]\green {\longmapsto\pmb {\sf x=-1} }[/tex]
[tex] \sf({ - 1)}^{3} - 6( { - 1)}^{2} - 2( - 1) + 12[/tex]
[tex] \sf - 1 - 6 + 2 + 12[/tex]
[tex]\sf 7≠0[/tex]
Therefore, x=-1 is not a zero.
[tex]\green {\longmapsto \pmb {\sf x=2}}[/tex]
[tex] \sf {(2)}^{3} - 6( {2)}^{2} - 2( 2) + 12[/tex]
[tex]\sf 8 - 24 - 4 + 12[/tex]
[tex]\sf -8≠0[/tex]
Therefore, x=2 is not a zero.
[tex]\green {\longmapsto \pmb{\sf x=-2 }}[/tex]
[tex]\sf (-2)^3-6(-2)^2 - 2( - 2) + 12[/tex]
[tex] \sf - 8 - 24 + 4 + 12[/tex]
[tex] \sf - 16≠0[/tex]
Therefore, x=-2 is not a zero.
[tex]\green {\longmapsto\pmb{\sf x=3}}[/tex]
[tex]\sf(3)^3-6(3)^2-2(3)+12[/tex]
[tex] \sf 27 - 54 - 6 + 12[/tex]
[tex]\sf -21≠0[/tex]
[tex]\green {\longmapsto \pmb {\sf x=-3}}[/tex]
[tex] \sf (-3)^3-6(-3)^2-2(-3)+12[/tex]
[tex] \sf - 27 - 54 + 6 + 12 [/tex]
[tex]\sf -63≠0[/tex]
Therefore, x=-3 is not a zero.
[tex]\green {\longmapsto \pmb {\sf x=4 }}[/tex]
[tex]\sf (4)^3-6(4)^2-2(4)+12[/tex]
[tex]\sf 64-96 - 8 + 12[/tex]
[tex]\sf -28≠0[/tex]
Therefore, x=4 is not a zero.
[tex]\green {\longmapsto \pmb{\sf x=-4 }}[/tex]
[tex]\sf (-4)^3-6(-4)^2-2(-4)+12[/tex]
[tex] \sf - 64 - 96 + 8 + 12[/tex]
[tex]\sf - 140≠0[/tex]
[tex]\green {\longmapsto \pmb{\sf x=6 }}[/tex]
[tex]\sf (6)^3-6(6)^2-2(6)+12[/tex]
[tex]\sf 216-216 - 12 + 12[/tex]
[tex]\sf 0≠0[/tex]
Therefore, x=6 is a zero.
[tex] \\ [/tex]
Now, dividing the polynomial by x-6.
[tex] \\ [/tex]
So, [tex]\sf x^2-2[/tex] is a factor of the polynomial.
[tex]\sf x^2-2[/tex]
[tex]\sf x=\pm \sqrt{2} [/tex]
[tex] \\ [/tex]
[tex]\large \pmb {\sf {\red {\bigstar Verification:-}}}[/tex]
Substituting x=6, ±√2 in the polynomial and solving :-
[tex] \\ [/tex]
_____________________
Hence, x=6, +√2, -√2 are the zeroes of the polynomial.
[tex] \\ [/tex]
Hope the concept's clear!
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
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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
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i Just Explained you the Method and the other hope it's understandable.