If P(x) and Q(x) have a common factor (x-

h),prove that (x-h) is also a factor of P(x)- Q
(x).find the possible common factors when P
(x)=ax^3 + x^2 - 15x - 18 and Q(x)= ax^3 - 14x -
12.also find the corresponding value of a.

Plz show me step of work

This just the same as showing that property with numbers.

If a|p and a|q then a|p-q

p = a*m and q=a*n
p-q = a(m-n) so a|p-q

for your polynomials,
P-Q = (ax^3+x^2-15x-18)-(ax^3-14x-12)
= x^2-x-6
= (x-3)(x+2)

that should get you going, right?

no sir plz i dont know it

huh? surely you recall the distributive property

(x-h) divides P(x), so P(x) = (x-h)*p(x)
similarly, Q(x) = (x-h)*q(x)

So, P-Q = (x-h)*p(x) - (x-h)*q(x)
= (x-h)(p(x)-q(x))
so, x-h divides P-Q

And above, I showed you what that is, and even factored it for you! So, those factors are the only possible common factors of P and Q.

As for finding a, you just need to find a so that those factors divide P and Q.

Sir please explain more.... help us in solving it more

To prove that (x-h) is also a factor of P(x)-Q(x), we need to show that when we divide P(x)-Q(x) by (x-h), the remainder will be zero.

1. First, let's find P(x)-Q(x):

P(x) = ax^3 + x^2 - 15x - 18
Q(x) = ax^3 - 14x - 12

P(x)-Q(x) = (ax^3 + x^2 - 15x - 18) - (ax^3 - 14x - 12)
= ax^3 + x^2 - 15x - 18 - ax^3 + 14x + 12
= x^2 - x + 6

2. Now, let's divide (x-h) into P(x)-Q(x) using polynomial long division:

x - 1
__________________
x - h | x^2 - x + 6
- (x^2 - hx)
___________
- hx + 6

Since the resulting remainder is -hx + 6, we can conclude that (x-h) is a factor if the remainder is zero.

To find the possible common factors, we need to set the remainder equal to zero:

-hx + 6 = 0

Solving for x, we get:
x = 6/h

For (x-h) to be a factor, x should equal h. Therefore, h = 6/h.

Simplifying the expression by multiplying both sides by h, we get:
h^2 = 6

To find the corresponding value of a, let's equate the coefficients of x^3 in P(x) and Q(x):

ax^3 = ax^3

Since these are equal, we can say that the coefficients of x^3 in both P(x) and Q(x) are the same, regardless of the value of 'a'.

In summary:
- (x-h) is a common factor of P(x) and Q(x) if the remainder of (x-h) divided into P(x)-Q(x) is zero.
- The possible common factors in this specific case are h = 6.
- The corresponding value of 'a' does not affect the common factors because the coefficients of x^3 in both P(x) and Q(x) are unaffected by 'a'.