Question

using factor theorem show that (x+√2) is a factor of (7x²-4√2x-6)​

Answer 1

If x+√2 is the factor of the given eqn,

then x = -√2 is the root of the solution.

f(-√2) = 7(√2)² - 4(√2)(√2) - 6

= 7(2)- 4(2)-6

=14 - 8 - 6

=14 - 14

= 0

Thus, x + √2 is a factor of the given eqn.

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Answer 2

Answer:

Step-by-step explanation:

$Zero \ of \ (x+\sqrt{2} )\\=>x =-\sqrt{2} \\Let \ P(x) = 7x^2-4\sqrt{2} x-6\\P(-\sqrt{2} )=7(-\sqrt{2} )^2 -4\sqrt{2} (-\sqrt{2} )-6\\P(-\sqrt{2} )=14+8-6\\P(-\sqrt{2} )=16\\Hence(x+\sqrt{2} )\ is \ not \ a \ factor \ of\ polynomial\ P(x)$

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