If (x – b) is a factor of p(x) = x2 + ax – 20, then show that b = 20/(a + b), where a + b ≠ 0.
Answers
Given,
(x – b) is a factor of p(x) = x² + ax - 20.
To prove,
b = 20/(a+b), where (a+b) ≠ 0
Solution,
We can easily solve this problem by following the given steps.
According to the question,
(x – b) is a factor of p(x) = x² + ax - 20
So, we have
(x-b) = 0
x = b
Now, putting the value of x in the given polynomial,
p(x) = x² + ax - 20
p(b) = b² + ab - 20
We know that the factor makes the value of the polynomial to be 0.
b² + ab - 20 = 0
b²+ab = 20 ( Moving 20 from the left-hand side to the right-hand side results in the change of the sign from minus to plus.)
Taking b common,
b(b+a) = 20
b = 20/(a+b) [ (a+b) was in the multiplication on the left-hand side. So, it is in the division on the right-hand side.]
Hence, it is proved that b = 20/(a+b).
Answer:
Given,
(x - b) is a factor of p(x) = x ^ 2 + ax - 20
To prove,
b = 20 / (a + b) where (a+b) = 0
Solution,
We can easily solve this problem by following the given steps.
According to the question, (x-b) is a factor of p(x) = x ^ 2 + ax - 20
So, we have
(x - b) = 0
x=b
Now, putting the value of x in the given polynomial,
p(x) = x ^ 2 + ax - 20
p(b) = b ^ 2 + ab - 20
We know that the factor makes the value of the polynomial to be 0.
b ^ 2 + ab - 20 = 0
b²+ab = 20 (Moving 20 from the left-hand side to the right-hand side results in the change of the sign from minus to plus.)
Taking b common,
b(b + a) = 20
b=20/(a+b) [(a+b) was in the multiplication on the left-hand side. So, it
is in the division on the right-hand side.]
Hence, it is proved that b = 20/(a+b).