divide: 2x^4-6x^3+3x^2+3x-2 by x+1 and x+2
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First we factorize divisor x² - 3x+2
= x² -2x - x +2 = x(x-2) -1 ( x - 2)
= ( x-2) (x-1)
Now, dividend p(x) = 2x^4 - 6x^3 + 3x² + 3x -2
If p(x) is dividible by ( x-2) & (x-1) each, then p(x) is also divisible by its product.
& by remainder theorem , if p(2) = 0 & p(1) =0
Then, p(x) is divisible by the divisor.
P( 2) = 2* 2^4 - 6* 2^3 + 3* 2² + 3*2 -2
= 32 - 48 + 12 + 6 - 2 = 50 - 50
= 0 ….. (1)
Also p(1) = 2* 1^4 - 6* 1^3 + 3 *1² + 3*1 -2
= 2 - 6 +3 + 3 - 2
= 0 ….. (2).
By (1) & (2), we can conclude that
P(x) is exactly dividible by x² - 3x + 2
Step-by-step explanation:
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