Question

question ⤵
Without actual division, prove that
$\sf {x}^{4} + 2 {x}^{3} - 2 {x}^{2} + 2x - 3$
Is exactly divisible by
$\sf {x}^{2} + 2x - 3.$
No spams needed.
Quality Answers needed.
A. S. A. P. ​

Answer 1

Step-by-step explanation:

suppose x⁴+2x³-2x²+2x-3 = f(x)

x²+2x-3 = 0

=> x²+(3-1)x-3 = 0

=> x²+3x-x-3 = 0

=> x(x+3)-1(x+3) = 0

=>(x+3)(x-1) = 0

=> x = -3 => x = 1

-3 and 1 are the roots of x²+2x-3 = 0

if x⁴+2x³-2x²+2x-3 is divisible by x²+2x-3 then f(-3) and f(1) = 0

f(-3) = (-3)⁴+(2*(-3)³)-(2*(-3)²)+(2*(-3))-3

= 81+(2*(-27))-(2*9)-6-3

= 81-54-18-6-3

= 81-81

= 0

f(1) = (1)⁴+(2*(1)³)-(2*(1)²)+(2*(1))-3

= 1+2-2+2-3

= 5-5

= 0

this is proved that x⁴+2x³-2x²+2x-3 is divisible by x²+2x-3