Question

x sq+2x-3is factor of f (x) x^4+6x^3+2ax^2 +bx-3a

Answer 1

This problem deals with many diverse applications in higher algebra.

Since x+a is a factor of given polynomial (f(x)), which implies polynomial can be written as,

f(x)=(x+a)*quotient(and no remainder as it is factor of f(x).) So at x=-a f(x) becomes zero.

Solving,

−a3+a3+3a+4=0−a3+a3+3a+4=0

a=−4/3a=−4/3

NOTE: This method is far more useful and short than long division method.

P.S.: I knew the answer straight away. The reason for writing a long answer is quora may collapse anwers.or......This is pretty simple. Using basics of remainder theorem, remainder of f(x) when divided with (x-a) will be f(a).

So from the above data, f(-1) = 6 and f(1) = 0.

Substituting it, 3(-1)^4-a(-1)^3+2a(-1)^2-(-1)-b = 6

i.e. 3+a+2a+1-b = 6 => 3a - b = 2

Also 3(1)^4-a(1)^3+2a(1)^2-(1)-b = 0

i.e. 3-a+2a-1-b = 0 => a - b = -2

Solving these two equations, we can

Answer 2

x²+2x-3 = (x+3)(x-1)
x+3=0, x= -3
x-1=0, x = -1
f(x) = x⁴+6x³+2ax²+bx-3a
f(-3) =(-3)⁴+6(-3)³+2a(-3)²+b(-3)-3a
0 = 81 - 162 + 18a - 3b -3a
15a - 3b = 81
5a - b = 27 -------> Eq 1
f(1) =(1)⁴+6(1)³+2a(1)²+b(1)-3a
0 = 1 + 6 +2a +b -3a
a - b = 7--------> Eq 2
From Eq1 & Eq2
5a - b = 27
a - b = 7
--------------------
4a = 20
a = 5
Sub a = 5 in Eq 2
5 - b = 7
b = -2