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Answer:
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Step-by-step explanation:
x+
2
) is a factor of (2\sqrt{2}x^2+5x+\sqrt{2})(2
2
x
2
+5x+
2
) by using remainder theorem.
Solution:
Let
f(x)=2\sqrt{2}x^2+5x+\sqrt{2}f(x)=2
2
x
2
+5x+
2
Using remainder theorem
Substitute
x+\sqrt{2}=0x+
2
=0
x=-\sqrt{2}x=−
2
Substitute the value of x in f(x)
f(-\sqrt{2})=2\sqrt{2}\times 2+5(-\sqrt{2})+\sqrt{2}f(−
2
)=2
2
×2+5(−
2
)+
2
f(-\sqrt{2})=4\sqrt{2}-5\sqrt{2}+\sqrt{2}f(−
2
)=4
2
−5
2
+
2
f(-\sqrt{2})=5\sqrt{2}-5\sqrt{2}=0f(−
2
)=5
2
−5
2
=0
Remainder=0
Therefore,x+\sqrt{2}x+
2
is a factor of2\sqrt{2}x^2+5x+\sqrt{2}2
2
x
2
+5x+
2
.
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Answer:
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