Question

if(x-4) is the HCF of (x^2+x--12) and (x^2-mx-8) then the value of m is:
a.0
b.1
c.2
d.6​

Answer 1

Answer:

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Answer 2

Compete question:

If (x-4) is the HCF of (x^2-x-12) and (x^2-mx-8) then the value of m is:

a.0

b.1

c.2

d.6​

Answer:

The correct answer is option(c) 2

Step-by-step explanation:

Given,

(x-4) is the HCF of (x²-x-12) and (x²-mx-8)

To find,

The value of 'm'

Solution:

Recall the theorem:

Factor theorem:

If (x-a) is a factor of the polynomial p(x), then p(a) = 0

Since (x-4) is the HCF of (x²-x-12) and (x²-mx-8) , we have (x-4) is a factor of  (x²-x-12) and (x²-mx-8)

Let p(x) = (x²-x-12) and q(x) = (x²-mx-8)

Hence we can say that (x-4)  is a factor of p(x) and q(x)

Since (x-4) is factor of q(x), then by factor theorem we have q(4) = 0

q(4) = 0 ⇒4²-4m-8 = 0

⇒16-4m-8 = 0

⇒8-4m= 0

⇒4m = 8

⇒m = 2

∴ The value of m = 2

Hence the correct answer is option(c) 2

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