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# Question 1 Determine which of the following polynomials has (x+1) a factor: (i) x3+x2+x+1 (ii) x4+x3+x2+x+1 (iii) x4+3x3+3x2+x+1 (iv) x3−x2−(2+√2)x+√2

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Solution

## To have (x + 1) as a factor, substituting x = - 1 must give p(-1) = 0. (i) x3+x2+x+1=(−1)3+(−1)2+(−1)+1=−1+1−1+1=0 Therefore, x+1 is a factor x3+x2+x+1. (ii) x4+x3+x2+x+1=(−1)4+(−1)3+(−1)2+(−1)+1=1−1+1−1+1=1 Remainder is not 0. Therefore (x + 1) is not its factor. (iii) x4+3x3+3x2+x+1=(−1)4+3(−1)3+3(−1)2+(−1)+1=1−3+3−1+1=1. Remainder is not 0. Therefore, (x + 1) is not a factor. (iv) x3+x2−(2+√2)x+√2=(−1)3−(−1)2−(2+√2)(−1)+√2=−1−1+2+√2+√2=2√2 Remainder not 0. Therefore, (x + 1) is not a factor.

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