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Tejaswini J answered this

the given polynomial isis divided by the other polynomial.

remainder = x+a .

since the degree of the polynomial is 4 and degree of the divisor is 2.

therefore the degree of the quotient must be 2.

let the quotient be

dividend = divisor * quotient + remainder

now equating the coefficients of different powers of x:

....................(1)

subtracting eq(2) from eq(1):

put the value of k = 5 in eq(2):

now equating the constant term:

thus k = 5 and a = - 5

hope this helps you.

cheers!!

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Tejaswini J answered this

Dear student,

We know that,

Dividend = Divisor × Quotient + Remainder

⇒ Dividend – Remainder = Divisor × Quotient

⇒ Dividend – Remainder is always divisible by the divisor.

Now, it is given that f(x) when divided by x²– 2x + k leaves (x + a) as remainder.

So, for f(x) to be completely divisible by x² – 2x + k, remainder must be equal to zero

⇒ (–10 + 2k)x + (10 – a – 8k + k²) = 0

⇒ –10 + 2k = 0 and 10 – a – 8k + k² = 0

⇒ k = 5 and 10 – a – 8 (5) + 5² = 0

⇒ k = 5 and – a – 5 = 0

⇒ k = 5 and a = –5

Regards.

Neha Thomas answered this

We know that,

Dividend = Divisor × Quotient + Remainder

⇒ Dividend – Remainder = Divisor × Quotient

⇒ Dividend – Remainder is always divisible by the divisor.

Now, it is given that f(x) when divided by x² – 2x + k leaves (x + a) as remainder.

So, for f(x) to be completely divisible by x² – 2x + k, remainder must be equal to zero

⇒ (–10 + 2k)x + (10 – a – 8k + k²) = 0

⇒ –10 + 2k = 0 and 10 – a – 8k + k²= 0

⇒ k = 5 and 10 – a – 8 (5) + 5² = 0

⇒ k = 5 and – a – 5 = 0

⇒ k = 5 and a = –5

-1

Neha Thomas answered this

We have,

$f (x) = x^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10$

Given that the remainder is x + a when f(x) is divided by x ² – 2x + k

i.e. we can write

$x^{4} - 6 x^{3} + 16 x^{2} - 25 x + 10 = (x^{2} - 2 x + k) � (x^{2} + b x + c) + (x + a) = x^{4} - 2 x^{3} + k x^{2} + b x^{3} - 2 b x^{2} + b k x + c x^{2} - 2 c x + c k + x + a = x^{4} + (b - 2) x^{3} + (k + c - 2 b) x^{2} + (b k - 2 c + 1) x + (c k + a)$

Comparing to coefficients of x ³, x ², x and constants on both sides, we get

and

ck + a = 10 ....(3)

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

k = 5

∴ From (1), c = 3 and from (3), we get a = –5

Hence k = 5 and a = –5

-1

Neha Thomas answered this

the given polynomial isis divided by the other polynomial.

remainder = x+a .

since the degree of the polynomial is 4 and degree of the divisor is 2.

therefore the degree of the quotient must be 2.

let the quotient be

dividend = divisor * quotient + remainder

now equating the coefficients of different powers of x:

....................(1)

subtracting eq(2) from eq(1):

put the value of k = 5 in eq(2):

now equating the constant term:

thus k = 5 and a = - 5

-1