Answer :

x^{3} – 6x^{2} + mx + 60 divided by (x + 2) and remainder = 2

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = x^{3} – 6x^{2} + mx + 60 and we have (x + 2)

The zero of (x + 2) is –2

Now using Remainder theorem,

p(x) = x^{3} – 6x^{2} + mx + 60 is divided by x + 2 then, p(–2) is the remainder which is 2

p(–2) = x^{3} – 6x^{2} + mx + 60 = 2

= (–2)^{3} – 6(–2)^{2} + m(–2) + 60 =2

= – 8 – 24 – 2m + 60 = 2

= – 32 – 2m + 60 = 2

= 28 – 2m = 2

= 2m = 28 – 2

= 2m = 26

m = 13

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