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?
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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
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