Question

If two polynomials x^2+px+q andx^2+mx+n has a common factor (x+a), then prove that, a=n-q/m-p

Answer 1

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$= > f(x) = {x}^{2} + px + q \\ \\ = > g(x) = {x}^{2} + mx + n$
If (x +a) is a factor of f(x) and g(x) , then,

f(x) =0 ; g(x) = 0____(when x = - a)

So,.
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$= > f(x) = {x}^{2} + px + q \\ \\ or. \: \: 0 = {( - a)}^{2} + p( - a) + q \\ \\ or. \: \: {a}^{2} - pa + q = 0 \\ \\ or. \: \: {a}^{2} = pa - q........(1)$
Again,
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$= > g(x) = {x}^{2} + mx + n \\ \\ or. \: \: 0 = {( - a)}^{2} + m( - a) + n \\ \\ or. \: \: {a}^{2} - ma + n = 0 \\ \\ or. \: \: {a}^{2} = ma - n.....(2)$
Comparing equation (1) and (2). We get
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$= > pa - q = ma - n \\ \\ or. \: \: ma - pa = n - q \\ \\ or. \: \: a(m - p) = (n - q) \\ \\ or. \: \: a = \frac{n - q}{m - p} .... < proved > ....$
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Answer 2

Answer:

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