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Question Number 157853 by emanuelMcCarthy last updated on 28/Oct/21

Commented by cortano last updated on 29/Oct/21
$(21)f(2)=−20 16+4p+16=−20 p+8=−5⇒p=−13 (22) f(−a)=−56 2a^3 −3a^2 −13ab+20=−56 2a^3 −3a^2 −13ab+76=0...(1) f(a)=0 −2a^3 −3a^2 +13ab+20=0...(2) (1)+(2)⇒−6a^2 =−96 ⇒a=± 4 (1)−(2)⇒4a^3 −26ab+56=0 ⇒b=((2a^3 +28)/(13a)) { ((a=4⇒b=((2.4^3 +28)/(13.4))=((32+7)/(13))=3)),((a=−4⇒b=((2(−4)^3 +28)/(13(−4)))=((32−7)/(13))=((25)/(13)))) :}$
$(21) f (2) = - 20$ $16 + 4 p + 16 = - 20$ $p + 8 = - 5 \Rightarrow p = - 13$ $(22) f (- a) = - 56$ $2 a^{3} - 3 a^{2} - 13 a b + 20 = - 56$ $2 a^{3} - 3 a^{2} - 13 a b + 76 = 0 . . . (1)$ $f (a) = 0$ $- 2 a^{3} - 3 a^{2} + 13 a b + 20 = 0 . . . (2)$ $(1) + (2) \Rightarrow - 6 a^{2} = - 96$ $\Rightarrow a = \pm 4$ $(1) - (2) \Rightarrow 4 a^{3} - 26 a b + 56 = 0$ $\Rightarrow b = \frac{2 a^{3} + 28}{13 a}$ ${\begin{cases} a = 4 \Rightarrow b = \frac{2 . 4^{3} + 28}{13.4} = \frac{32 + 7}{13} = 3 \\ a = - 4 \Rightarrow b = \frac{2 {(- 4)}^{3} + 28}{13 (- 4)} = \frac{32 - 7}{13} = \frac{25}{13} \end{cases}$
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