Question

100 points!! ASAP!!

28th and 29th

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Answer 1

Question :-

$If\:(x + a)\: is\: factor\: of\: two \: polynomials\: x^{2} + px + q\: and\\ x^{2} + mx + n,\:then\:prove\:that\: a =\:\frac{n - q}{m - p}.$

$\underline\bold{Answer}$

$(x + a)\: is\:the\:factor\:of\:the\:given\:polynomial\: {x}^{2} + px + q$

Therefore,

$(-{a}) ^{2} + p(-a) + q = 0 \\ {a}^{2}- ap + q = 0 - - (1) \\ (x + a) \: is \: the \: factor \: of \: {x}^{2} + mx + n \\ ( -a)^{2} + m( - a) + n = 0 \\ {a}^{2} - am + n = 0 - - (2)$

From equation (1) and (2) we get,

-ap + am + q -n = 0

a (m - p) = n - q [ Taking a common ]

$\large\bold{a = \frac{n - q}{m - p}}$

*Refer the attachment for Second question Answer,

Step 1 = Let us consider p(x) as quadratic equation

Step 2 = The sums of roots α + β = - b/a

Step 3 = The product of roots αβ = = c/a

Over here,

a = a + 1

b = 2a + 3

c = 3a + 4

Answer 2

Hope attachment can help you!