Question

If p(x)=x2−kx+8 [3]
a) Find p (2)?
b) What is the value of k if x−2 is a factor of p(x)?
c) Write p(x) as the product of two first degree polynomials if one of its factors is x−2?

Answer 1

Answer:

p(x)=x

2

+6x+k

(a) If k = 10

p(x)=x

2

+6x+10

Discriminant =b

2

−4ac=6

2

−4(1)(10)=36−40=−4

Since the discriminant is negative, this equation has no solution. Hence, p(x) has no first degree factors.

(b) p(x)=x

2

+6x+k

For p(x) to have first degree factor, the discriminant of the equation x

2

+6x+k=0

should be zero or positive.

i.e. b

2

−4ac=6

2

−4k≥0

i.e. 6

2

≥4k

i.e.

4

36

≥k

i.e. k≤9

Thus, the maximum value of k is 9.

(c) If k = -1, then p(x) =x

2

+6x−

Step-by-step explanation:

follow the prachi vala