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Given that (x^2−4) is a factor of the polynomial  f(x), where f(x) =4x^4+7x^3 +ax^2 +bx +8. Find the values of a and b and hence factorize f(x)completley. Find the set of values for which f(x)=0

 Dec 2, 2017
 #1
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If x^2 - 4  is a factor.....then  (x - 2)  and (x + 2)  are  factors

Which means that   2  and - 2 are roots

 

So

4(2)^4+7(2)^3 +a(2)^2 +b(2) +8  = 0

4(-2)^4+7(-2)^3 +a(-2)^2 + b(-2) +8  = 0

 

Add these and we have that

 

128   +  8a   +   16  = 0    ⇒  8a  =  -144  ⇒   a  =  - 18

 

Subtract them and we have that

 

112 + 4b   =   0

28 + b  =  0    ⇒   b  = -28 

 

So....the polynomial is

 

4x^2  + 7x^3  - 18x^2 - 28x  + 8       and we can write this as

 

[4x^2  + 7x^3  - 2x^2]  -  16x^2  - 28x + 8  

x^2 (4x^2  + 7x - 2)  -  4 ( 4x^2 + 7x - 2 )

(x^2 - 4) (4x^2 + 7x - 2)

(x^2 - 4) (4x - 1) (x + 2) 

(x - 2) (x + 2) (4x - 1) (x + 2)  =  0

 

The  values that make this  = 0   are     x= 2, x = 1/4  and x  = -2

 

 

cool cool cool

 Dec 2, 2017

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