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secondary 3 | A Maths
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binomial
Date Posted: 4 years ago
Views: 270
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We need to expand a few terms out and collect all the relevant terms in x^6 and x^5.
In (2x + q)..., the 2x must be multiplied by some x^5 to get a term in x^6 and by some x^4 to get a term in x^5, while the q must be multiplied by some x^6 to get a term in x^6 and by some x^5 to get a term in x^5.
From there, we collect all the terms in x^6 and equate its coefficient to -7, and all the terms in x^5 and equate its coefficient to 0. The rest is simultaneous equation solving.
In (2x + q)..., the 2x must be multiplied by some x^5 to get a term in x^6 and by some x^4 to get a term in x^5, while the q must be multiplied by some x^6 to get a term in x^6 and by some x^5 to get a term in x^5.
From there, we collect all the terms in x^6 and equate its coefficient to -7, and all the terms in x^5 and equate its coefficient to 0. The rest is simultaneous equation solving.
Date Posted:
4 years ago
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