Polynomials Test - 4

If the polynomial x⁴ - 6x³ + 16x² - 25x + 10 on division by x²  - 2x + k leaves remainder (x + a),

then, polynomial (x⁴ - 6x³ + 16x² - 25x + 10) - (x + a) on division by x²  - 2x + k leaves remainder zero.

Using long division we get,

(x⁴ - 6x³ + 16x² - 25x + 10) - (x + a) = x⁴ - 6x³ + 16x² - 26x + 10 - a by x²  - 2x + k ,

the remainder obtained is (-10 + 2k ) x + ( 10 - a - 8k + k²) 

So, (-10 + 2k ) x + ( 10 - a - 8k + k²) = 0

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

⇒   k = 5  and a = - 5

A polynomial of degree 2 is called a quadratic polynomial.

The polynomials having -2 and 5 as the zeroes can be written in the form

k(x + 2) (x - 5), where k is a constant.

Thus, number of polynomials with roots -2 and 5 are infinitely many, since k can take infinitely many values.

P(a) = x²a² - 2xa + 2x - 4

Since 1 is a zero, so P(1) = 0

⇒ P(1) = x²1² - 2x.1 + 2 x - 4 = 0

x² - 4 =0

⇒ x =±2

The process of division is stopped when the remainder is zero or its degree is less than the degree of the divisor.

Hence, if the degree of the divisor is one, the degree of the remainder has to be 0.

If (x + 6) is a factor of the given polynomial, then the remainder should be zero. 

Dividing the polynomial by (x + 6) and equating the remainder, (a – 12) to 0.

Hence, a – 12 = 0

Or, a = 12.

Recall,

f(x) = g(x) q(x) + r(x)

To get g(x), divide f(x) – r(x) by q(x).

Hence, g(x) = x² - x + 1.

The exponent or the highest power of the variable that occurs in a polynomial, in this case 3, is called the degree of the polynomial.

(5,0) is the point, where the graph cuts the X- axis. Therefore, x = 5 is the zero of the polynomial.