# Create a polynomial of degree 2 having a remainder of 9 when it is divided by (x+9).

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### 2 Answers

A polynomial of degree 2 has to be created that has a remainder of 9 when it is divided by (x+9).

There are several such polynomials that can be created. It has to be of the form (x + a)(x + 9) + 9 where a is a real number

=> x^2 + (9+a)x + 9a + 9

**A polynomial x^2 + (9+a)x + 9a + 9 where a is a real number has a remainder 9 when divided by (x+9)**

You need to remember that reminder theorem tells you that if you divide a polynomial by a linear binomial x-a, the reminder is a constant value, namely f(a)=r(a).

You should write the reminder theorem such that:

`ax^2 + bx + c = (x + 9)(x + d) + 9`

Notice that quotient is x + d and reminder is 9.

Opening the brackets yields:

`ax^2 + bx + c = x^2 + x(d+9) + 9`

Equating the coefficients of like powers yields:

a=1

b = d+9

c = 9

Using the reminder theorem yields that f(-9)=9, hence, you need to substitute -9 for x in equation of quadratic such that:

`f(-9)=9 =gt 81 - 9b + 9 = 9 =gt 9b= 81 =gt b = 9`

**Hence, evaluating the quadratic polynomial under the given conditions yields `f(x) =x^2 + 9x + 9` .**