JEE MATHEMATICS MCQ: THEORY OF EQUATION

1 Easy

1 of 1

Application +4 | -1

x + 1

is a factor of

{x^4} + \left( {p - 3} \right){x^3} - \left( {3p - 5} \right){x^2} + \left( {2p - 9} \right)x + 6,

then the value of

p

View answer and explanation

Correct option: C

Option C (4) is correct because substituting x = -1 into the polynomial and setting the result to zero allows us to solve for p, which equals 4.

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{x^{2n}} - 1 = 0

\frac{1}{x}+\frac{1}{\sqrt{\left(1-x^2\right)}}=\frac{35}{12}

x^7+14 x^5+16 x^3+30 x-560=0

x_1, x_2, x_3, \ldots, x_n

x^n+a x+b=0

\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right) \ldots\left(x_1-x_n\right)

P(x)=k x^3+3 x^2-3

Q(x)=2 x^3-5 x+k

(x-4)

k

{\left( {x - 1} \right)^3}

{x^4} + a{x^3} + b{x^2} + cx - 1,