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If P(x) = ax4+bx3+cx2+dx+e has roots at x = 1, 2, 3, 4 and P (0) = 48, what is P(5)

If P(x) = ax4+bx3+cx2+dx+e has roots at x = 1, 2, 3, 4 and P (0) = 48, what is P(5)

Grade:9

1 Answers

Arun
25763 Points
3 years ago
that means we can write the polynomial as-
a(x –1)(x –2)(x –3)(x –4) = P(x)
now P(0) = 48
hence
a *( –1)( –2)( –3)( –4) = 48
hence a = 48/24 = 2
now p(x) = 2 (x –1)(x –2)(x –3)(x –4)
hence P(5) = 2* 4* 3* 2* 1
P(5) = 48
hope it helps

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