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Is the set of real polynomials (with the degree of 7 and less, and with the roots 2 and 3) a vector space? If yes, find its basis.

So we can write that polynomial as: $p(x) = (x-2)*(x-3)*(ax^5+bx^4+cx^3+dx^2+ex+f)$

Yes, it is a vector space, because the degree of our polynomial would be preserved if we add another polynomial or multiply it by a scalar and I understand how to prove it, but what about that basis? Should I multiply those brackets?

And if we have this linear mapping $(\mathbb R^7\to\mathbb R^8)$ : $p_1(x) \to p_2(x)=(x-2)*(x-3)*(ax^5+bx^4+cx^3+dx^2+ex+f)$ , how would look like the Kernel and the Image? Let say that $p_1(x)$ would be a standard polynomial with the degree of 6. I made up this example, so maybe I forgot something or wrote it wrongly.

I would let $p_2(x)$ = 0 if I want to find Kernel but how would look like it properly? The image is just that $p_2(x)$, or $p_2(x) = y$, if this equality makes sense?

Thank you for the help.

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Hint with basis: in the factorization of $p(x)$ the term $ax^5+\dots$ could be any polynomial of degree $\le 5$.

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    Can I use just canonical basis {$1, x, x^2,...,x^7$}? – 2017-01-30
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    Are you dealing with any polynomial of degree at most 7? I don't think so. Use the canonical basis for polynomials of degree at most 5 and do something with it. – 2017-01-30
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    I don't understand, I thought I am working with the polynomial of degree at most 7, just that I know only 2 roots. – 2017-01-30
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    So, only 5 roots are free, hence your subspace is 6-dimensional. Does, for instance, $x(x^6+1)$ belong to this subspace? – 2017-01-30
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    {$x^5(x-2)(x-3), x^4(x-2)(x-3), ..., x(x-2)(x-3),(x-2)(x-3)$} – 2017-01-30
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    Now it is quite OK – 2017-01-30