Question

Please solve it very urgent .
Find a & b so that x4 + ax³ -7x² - 8x + b is exactly divisible by ( x + 2 ) & ( x + 3 ).

Answer 1

x⁴+ax³-7x²-8x+b

(x+2) and (x+3) are factors of it

x+2 = 0 ; x+3 = 0

x = -2 ; x = -3

Put x= -2 and-3 to fund a and b

First put x= -2

(-2)⁴+a(-2)³-7(-2)²-8(-2)+b = 0

16+a(-8)-7(4)+16+b=0

32-8a-28+b = 0

-8a+b+4 = 0

-8a+b = -4 ---------(1)

Put x= -3

(-3)⁴+a(-3)³-7(-3)²-8(-3)+b = 0

81+a(-27)-7(9)+24+b = 0

81-27a-63+24+b = 0

-27a+b+42 = 0

-27a+b = -42 ------(2)

(2) - (1)

-27a+b = -42
-8a + b = -4
-----------------
-19a = -38

19a = 38

a = 38/19 = 2

Put a=2 in (1)

-8(2)+b= -4

-16+b = -4

b = -4+16

b = 12

Therefore, a = 2 and b = 12

Hope it helps

Answer 2

x⁴+ax³-7x²-8x+b
(x+2) and (x+3) are factors of the given expression,
x = -2&-3
Putting x=-2,we get
(-2)⁴+a(-2)³-7(-2)²-8(-2)+b = 0
=32-8a-28+b
=-8a+b+4
-8a+b = -4 ................(i)
Put x= -3,we get,
(-3)⁴+a(-3)³-7(-3)²-8(-3)+b = 0
=81-27a-63+24+b
=-27a+b+42
-27a+b = -42..............(ii)
Subtracting (i) from (ii),we get,
-19a = -38
19a = 38
a= 2

now,

Put a=2 in(i),
-8(2)+b= -4
b = -4+16
b = 12