Question

(x-√2) is a factor of (7x²-4√2x-6)​

Answer 1

Answer:

let p(x)= 7x^2-4root2x-6

By,the factor theorem ,

g(x)= x-root 2 will be a factor if p(root2)=0

$x - \sqrt{2} \: will \: be \: a \: factor \: of \: p(x) \: if \: p( \sqrt{2} ) = 0 \\ p( \sqrt{2} ) = 7( { \sqrt{2} }^{} ) {}^{2} - 4 \sqrt{2} ( \sqrt{2} ) - 6 \\ \: \: \: \: \: \: \: = 14 - 8 - 6 \\ \: \: \: \: \: \: = 14 - 14 = 0$

since p(root2)=0 ,

therefore g(x)is a factor of p(x)