Consider p(x)=3x^3+ax+5a, a is an element of the integers.
The polynomial p(x) leaves a remainder of -7 when divided by (x-a). Show that only one value of a satisfies the above condition and state it’s value.

Since you know a is an integer, you can try to factorize this, or use the rational root theorem to generate a list of potential rational roots, and from there you'd select the integer roots. The rational ones are of the form m/n, where m divides the constant term 7 and n divides the leading coefficient 3. They are

a = ± 1 or ± 1/3 or ± 7 or ±7/3

so there are 4 integer candidates,

a = -1 or 1 or -7 or 7

Check the value of f(a) = 3a³ + a² + 5a + 7 at each of these:

a = -1 → f(a) = 0

a = 1 → f(a) = 16

a = -7 → f(a) = -1008

a = 7 → f(a) = 1120

and only a = -1 is a root of 3a³ + a² + 5a + 7, so that's your answer. In fact,

3a³ + a² + 5a + 7 = (a + 1) (3a² - 2a + 7)

Consider p(x)=3x^3+ax+5a, a is an element of the integers. The polynomial p(x) leaves a remainder of -7 when divided by (x-a). Show that only one value of a satisfies the above condition and state it’s value.

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Consider p(x)=3x^3+ax+5a, a is an element of the integers.
The polynomial p(x) leaves a remainder of -7 when divided by (x-a). Show that only one value of a satisfies the above condition and state it’s value.