find the zero of polynomial
[tex] x^{3}-6x^{2}+11x-6[/tex]
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find the zero of polynomial
[tex] x^{3}-6x^{2}+11x-6[/tex]
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Given polynomial is [tex]x^{3} -6x^{2} +11x-6[/tex]
By using Trial and error method.
Put [tex]x=1[/tex]
[tex]1^{3} -6(1)^{2} +11(1)-6=0\\1-6+11-6=0\\12-12=0\\0=0[/tex]
[tex]1[/tex] is the root of equation. So [tex](x-1)[/tex] is a factor of [tex]x^{3} -6x^{2} +11x-6[/tex].
put [tex]x=2[/tex]
[tex]2^{3} -6(2)^{2} +11(2)-6=0\\8-6(4)+22-6=0\\8-24+22-6=0\\30-30=0\\0=0[/tex]
[tex]2[/tex] is a root of equation. So, [tex](x-2)[/tex] is a factor of [tex]x^{3} -6x^{2} +11x-6[/tex].
Put [tex]x=3[/tex]
[tex]3^{3} -6(3)^{2} +11(3)-6=0\\27-6(9)+33-6=0\\27-54+33-6=0\\60-60=0\\0=0[/tex]
[tex]3[/tex] is a root of equation. So, [tex](x-3)[/tex] is a factor of [tex]x^{3} -6x^{2} +11x-6[/tex].
Therefore, [tex]1,2,3[/tex] are the zeroes of the polynomial [tex]x^{3} -6x^{2} +11x-6[/tex].
Given,
[tex]x^{3}-6x^{2} +11x-6=0[/tex]
Substitute [tex]x=1[/tex] in the given equation,
[tex]1^{3}-6.1^{2}+11.1-6\\1-6+11-6\\12-12=0\\0=0[/tex]
[tex]x-1[/tex] is the one of the factor of [tex]x^{3}-6x^{2} +11x-6[/tex]
Now,
Divide the given equation with [tex](x-1)[/tex]
We, get quotient [tex]x^{2} -5x+6[/tex].
By using factorization splitting method to find the factors
[tex]x^{2} -5x+6=0[/tex]
[tex]x^{2} -3x-2x+6=0\\x(x-3)-2(x-3)=0\\(x-3)(x-2)=0\\x=2,3[/tex]
Therefore, [tex]1,2,3[/tex] are the zero polynomial of [tex]x^{3}-6x^{2}+11x-6[/tex].