Question

If x+a is the hcf of x²+px+q and x²+p'x+q' find a​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

↝ x + a is the hcf of x² + px + q and x² + p'x + q'

So,

It means x + a is the factor of x² + px + q and x² + p'x + q'

We know,

Factor theorem states that if x - a is a factor of polynomial f(x) then f(a) = 0.

Thus,

$\rm :\longmapsto\: {a}^{2} + pa + q \: = \: 0 - - - - (1)$

and

$\rm :\longmapsto\: {a}^{2} + p'a + q' \: = \: 0 - - - - (2)$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\:pa - p'a \: + \: q - q' \: = \: 0$

$\rm :\longmapsto\:(p - p')a \: = \: q' - q$

$\bf\implies \:a \: = \: \dfrac{q' - q}{p - p'}$

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Additional Information :-

Remainder Theorem states that if a polynomial f(x) is divided by linear polynomial x - a, the remainder is f(a).

More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

Answer:

a = p'−p/q'−q

Step-by-step explanation:

We have,

(x+a) is the HCF of x²+px+q and x²+p'x+q'

Put x=−a

Therefore,

⇒(−a)²+p(−a)+q=0

⇒a²−pa+q=0

⇒a² =pa−q             ............(1)

And

⇒(−a)²+p'(−a)+q'=0

⇒a²−p'a+q'=0

⇒a² =p'a−q'             ............(2)

equations (1) and (2), we get

p'a−q'=pa−q

a(p'−p)=q'−q

a= p'−p/q'−q

Therefore, the value of x is

x= p−p'/q−q'

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