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. Factorize f(x).?

Since (x^2−4) is a factor of the polynomial f(x) = 4x^4+7x^3+ax^2 +bx +8

Then by factor theorem x²-4=0 i.e. x= 2 or -2

f(2)= 4×2^4+7×2^3+a×2^2 +b×2 +8 =128+4a+2b=0................(i)

f(-2)= 4×(-2)^4+7×(-2)^3+a×(-2)^2 +b×(-2) +8 =16+4a-2b=0....(ii)

Adding  (i) and (ii) gives 144+8a=0 i.e. a=-144/8=-18

Subtracting (ii) from (i) gives 112+4b=0 i.e. a=-112/4=-28

f(x) = 4x⁴+7x³+ax²+bx+8

x²-4 = 0

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

x=2, -2

f(2) = 4(2)⁴+7(2)³+a(2)²+b(2) + 8

= 64 + 56 + 4a+2b + 8

= 128 + 4a + 2b = 0 ………..(1)

f(-2) = 64-56 + 4a – 2b + 8

= 16+4a-2b = 0 ……….(2)

From equation (1) and (2)

2a+b = -64

2a-b = -16

________

4a = -80

a = -20

b = -64 – 2a

= -64 + 40

b = -24

f(x) = 4x⁴+7x³-20x²-24x+8 = 0

(x² - 4) = (x - 2)(x + 2), so -2 and +2 are both factors of f(x).  This means if we divide f(x) by (x - 2) or (x + 2), we should get a zero remainder.

4......7...... a......b...... 8 ....... .....| -2

......-8 ......2....-2a-4 ..-2b+8+4a

4.....-1 ...a+2..b-4-2a .. -2b+16+4a and this must equal zero, so:

4a - 2b = -16

2a - b = -8 (*)

4......7...... a......b...... ..... 8 ....... .....| 2

........8......30 ...60+2a......2b+4a+120

4 ....15...30+a...b+2a+60...2b+4a+128 and this must equal zero too, so:

4a + 2b = -128

2a + b = -64 (**)

Add (*) and (**) together and get 4a = -72, or a = -18

b = -64 + 36 = -28

Thus, f(x) = 4x^4 + 7x³ - 18x² - 28x + 8 =

(x + 2)²(x - 2)(4x - 1)

 (x^2−4)=(x-2)(x+2) is a factor of the polynomial f(x)

---> Factor theorem

f(2)=0 and

f(-2)=0

solve 2 equations for a , b