Math, asked by anaagupta591, 2 months ago

If 3x+1 is a factor of the polynomial p(x) = 27x³ – ax² – x + 3, then find the value of ‘a’.

Answers

Answered by ks78182k
1

Answer:

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Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′Share

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswer

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr app

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option is

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option isA

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option isA21

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option isA21x=−31 is a zero of polynomial p(x)

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option isA21x=−31 is a zero of polynomial p(x)⟹p(−31)=0

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option isA21x=−31 is a zero of polynomial p(x)⟹p(−31)=0⟹27(−31)3−a(−31)2−(−31)+3=0

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option isA21x=−31 is a zero of polynomial p(x)⟹p(−31)=0⟹27(−31)3−a(−31)2−(−31)+3=0⟹−2727−9a+31+3=0

Join / LoginMathsIf x=−31 is a zero of a polynomial p(x)=27x3−ax2−x+3, then find the value of ′a′ShareAnswerOpen in answr appCorrect option isA21x=−31 is a zero of polynomial p(x)⟹p(−31)=0⟹27(−31)3−a(−31)2−(−31)+3=0⟹−2727−9a+31+3=0⟹9a=2+31

Answered by jn813782
2

Answer:

x=−

3

1

is a zero of polynomial p(x)

⟹p(−

3

1

)=0

⟹27(−

3

1

)

3

−a(−

3

1

)

2

−(−

3

1

)+3=0

⟹−

27

27

9

a

+

3

1

+3=0

9

a

=2+

3

1

9

a

=

3

7

⟹a=21

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