Question

On dividing x3-3x^2+5x-3 by a polynomial p(x) the quotient and remainder are x-3 and 7x-9 find p(x)

Answer 1

ANSWER

CONCEPT

$dividend = divisor × quotient \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + remainder \:$

SOLUTION

We have to find value of divisor

given

dividend = x3-3x^2+5x-3

divisor = p(x)

quotient = (x-3)

remainder = 7x-9

Put value of given quantities

x3-3x^2+5x-3 = p(x) × (x-3) + 7x-9

p(x) = x3-3x^2+5x-3 -7x+9

x-3

p(x) = x3-3x^2-2x+6

x-3

since ( x-3) is a factor of x3-3x^2-2x+6

on dividing we get

p(x) = x^2 -2

Answer 2

Question:

On dividing x³-3x²+5x-3 by a polynomial p(x) the quotient and remainder are x-3 and 7x-9 find p(x).

Solution:

Here,

Divisor = p(x)

Let the dividend, quotient and remainder be g(x) , q(x) and r(x) respectively.

By Division Algorithm,

Dividend = Divisor x quotient + Remainder

g(x) = p(x) × q(x) + r(x)

We have,

g(x) [dividend] = x³ - 3x² +5x -3

q(x) [quotient] = x - 3

r(x) [remainder] = 7x-9

p(x) [divisor] = ?

Now,

Putting values

x³ - 3x² +5x -3 = p(x) ×( x - 3) + 7x - 9

x³ - 3x² +5x -3 - (7x-9) = p(x) x (x-3)

x³ - 3x² +5x -3 -7x +9 = p(x) × (x-3)

x³ - 3x² -2x + 6 = p(x) × (x -3)

x³ - 3x² -2x +6/(x-3) = p(x)

Now dividing x³ -3x² -2x +6 by x-3

Here,

x² - 2

x-3 | x³ -3x² -2x +6

x³ -3x²

- + (both are cancelled)

-2x + 6

-2 x + 6

+ - (again both cancelled)

0 <= remainder

Hence, p(x) = x² - 2

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