Show thar R(x)= Q(x) + 5 is a square of first degree binomial

Question

Given the polynomial: Q(x)=x²-14x+ 5

Show that R(x)= Q(x)+5 is a square of first degree binomial.

Answer 1

Q(x) = x² - 14x + 5

R(x) = Q(x) + 5 = (x² - 14x + 5) + 5

R(x) = x² - 14x + 10

R(x) is not the square of a first degree binomial. To be a square of a first degree binomial, its discriminant would have to be zero.

D = b² - 4ac = (-14)² - 4(1)(10) = 156 ≠ 0

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