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# given that (x-1) and (x+1) are factors of px^3 + qx^2 - 3x - 7 find the values of p and q.

it is yet another a factorising polynomial sum.how could i find p n q

If we multiply (x-1)(x+1) = (x2-1) we now have to find what factor when multiplied by (x2-1) will yield px3+qx2-3x-7.  we know that the factor will have +7 as the only way that we can get -7 will be multiplying (-1)7, and the same way we know that in order to get px3 is if we have x2*px.  Now let is build the factor and test the equation.

(x2-1)(px+7) = px3 +7x2-px-7,  and we look at the original equation px3+qx2- 3x - 7 and we can conclude that q=7 and p=3 .

Nice!
Hi Rodela;
(x-1) and (x+1)=x2-1
px^3 + qx^2 - 3x - 7
(x2-1)
For the FOIL...
FIRST must be (x2)(px)=px3
OUTER must be (x2)(q)=qx2
INNER must be (-1)(3x)=-3x
LAST must be (-1)(7)=-7
(x2-1)(3x+7)
p=3
q=7

hello vivian!!!
Hi Rodela!
p X^3 + q X^2 - 3X - 7

Has a factor of ( X -1) , then X =1 is the root of the polynomial

p ( 1) ^3 + q ( 1) ^2 - 3 ( 1) - 7 =0

P + q = 10     ( 1)

Factor of  ( X +1) , means that X = -1 is the root of the polynomial:

p ( -1) ^3 + q(-1)^2 - 3 ( -1) -7 =0

-p + q = 4      ( 2)

From equation ( 1) & ( 2) :

2q = 14      q= 7

P + 7 = 10         p = 3

Test Polynomial :

3 X^3 + 7 X^2 - 3x  -7

3 X^2 - 3X + 7 X^2 -7

3 X^3- 3X  + 7 X^2 -7

3x ( X^2 - 1) + 7 ( X^2 -1) =

( X +1) ( X-1) ( 3X +7 ) = 0

Zeros : X = -1  , X = 1  , X = -7/3