P(x) is a second degree polynomial. One factor of P(x) is (x – 1) and if P(7)=0
then,
a) Write P(x) as the product of two first degree polynomials.
b) Find the values of x for which P(x) = 0
Answers
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−1
Consider x
2
+6x−1=0
Discriminant =b
2
−4ac=6
2
−4(1)(−1)=36+4=40
Since the discriminant is positive, p(x) can be written as the product of two first degree polynomials.
x=
2a
−b±
b
2
−4ac
=
2
−6±40
=
2
−6+
40
,
2
−6−
40
=
2
−6+2
10
,
2
−6−2
10
=−3+
10
,−3−
10
∴x
2
+6x−1=(x−(−3+
10
))(x−(−3−
10
))
=(x+3−
10
)(x+3+
10
)
(d) Let k=−m⇒p(x)=x
2
+6x−m
In the second degree equation x
2
+6x−m=0
Discriminant =b
2
−4ac=6
2
−4(1)(−m)=36+4m
Since this is positive, p(x) has two distinct first degree factors.