For the polynomial below, -1 is a zero.
h(x) = x^3 + 5x^2 + 5x + 1
Express h(x) as a product of linear factors.
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h(x) = x^3 - 5x^2 + 5x -1 ; 1 is a zero express h(x) as a linear product of two factors?
You cannot express a cubic polynomial as a linear function and there is no such thing as a linear product of factors. However, 1 is indeed a zero of h(x) (I checked) and this means that (x-1) is a factor of h(x). So you can divide h(x) by (x-1) and get a polynomial with no remainder. Use synthetic (or long division) to get: 1 | 1 -5 .5 -1 ........ 1 -4 1 ------------------- ....1..-4..1 == so h(x)=(x-1)(x²-4x+1) Hopefully this is what you need.
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