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Question Number 162787 by amin96 last updated on 01/Jan/22

happy new year

{a; b; c} \in Z - {0}

p (x) = {a x}^{2} + b x + c

p (a) = 0

p (b) = 0

p (1) = ?

Answered by alephzero last updated on 01/Jan/22

p (a) = {a a}^{2} + b a + c = a^{3} + b a + c = 0

p (b) = {a b}^{2} + b b + c = {a b}^{2} + b^{2} + c = 0

p (1) = a^{2} + b + c = ?

{\begin{cases} a^{3} + b a + c = 0 \\ {a b}^{2} + b^{2} + c = 0 \end{cases}

\Rightarrow {\begin{cases} a^{3} + b a = - c \\ {a b}^{2} + b^{2} = - c \end{cases}

\Rightarrow a^{3} + b a = {a b}^{2} + b^{2}

\Rightarrow a (a^{2} + b) = b^{2} (a + 1)

(a_{1}; b_{1}) = (- \frac{1}{2}; - \frac{1}{2})

(a_{2}; b_{2}) = (- 2; 4)

(a_{3}; b_{3}) = (0; 0)

But {a; b; c} \in Z - {0}

\Rightarrow (a; b) = (- 2; 4)

\Rightarrow {\begin{cases} - 2^{3} + (- 2) \times 4 = - c \\ - 2 \times 4^{2} + 4^{2} = - c \end{cases}

{\begin{cases} - 8 + (- 8) = - c \\ - 36 + 16 = - c \end{cases}

\Rightarrow - 16 = - c

\Rightarrow c = 16

a^{2} + b + c = {(- 2)}^{2} + 4 + 16 = 4 + 4 +

+ 16 = 8 + 16 = 24

p (1) = 24

Commented by mr W last updated on 01/Jan/22

s o l u t i o n i s n o t u n i q u e .

f (x) = x^{2} + x - 2 i s a l s o o k .

f (1) = 0 .

i n f a c t a n y a = b = n, c = - n^{2} (n + 1) w i t h

n \in Z i s o k .

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