`h(x)=x^3+3x^2+x-5`.
- Show `h(1)=0` (1 mark)
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- Express `h(x)` in the form `h(x)=(x-1)*g(x)` where `g(x)` is a quadratic factor. (2 marks)
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- Justify that `h(x)` only has one zero. (2 marks)
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Show Worked Solution
i. `h(x)=x^3+3x^2+x-5`.
`h(1) = 1+3+1-5=0`
ii. `h(x)=(x-1)*g(x)`
`text{By long division:}`
`h(x)=(x-1)(x^2+4x+5)`
iii. `text{Consider the roots of}\ \ y=x^2+4x+5`
`Δ = b^2-4ac=4^2-4*1*5=-4<0`
`text{Since}\ \ Δ<0\ \ =>\ \ text{No zeros (roots)}`
`:. h(x)\ text{only has 1 zero at}\ x=1\ (h(1)=0)`