Expand the expression (the result will be a cubic expression).
Author: shiroeExecutions: 7
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The default value for P in the input field is set to 1.
Enter P, a, b, and c (please do not leave any fields blank).
First, it outputs j3, j2, j1, and j0 in a form where P is not distributed.
It then distributes P and outputs k3, k2, k1, and k0 as well.
Left-hand side → Before distribution → After distribution.
P(x+a)(x+b)(x+c) = P(j3x^3 + j2x^2 + j1x + j0) = k3x^3 + k2x^2 + k1x + k0
Enter P, a, b, and c (please do not leave any fields blank).
First, it outputs j3, j2, j1, and j0 in a form where P is not distributed.
It then distributes P and outputs k3, k2, k1, and k0 as well.
Left-hand side → Before distribution → After distribution.
P(x+a)(x+b)(x+c) = P(j3x^3 + j2x^2 + j1x + j0) = k3x^3 + k2x^2 + k1x + k0
| j3 | (Before distribution) |
| j2 | (Before distribution) |
| j1 | (Before distribution) |
| j0 | (Before distribution) |
| k3 | (After distribution) |
| k2 | (After distribution) |
| k1 | (After distribution) |
| k0 | (After distribution) |
Digit display setting: 50 digits. Digit count can be changed. Complex number calculation mode.
Please be careful not to omit the negative sign.
Please pay attention to the imaginary unit i in the output field.
Please be careful not to omit the negative sign.
Please pay attention to the imaginary unit i in the output field.
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In this case, the result will be a cubic expression.
The form of the expression is
P(x+a)(x+b)(x+c) = p(j3x^3 + j2x^2 + j1x + j0)
Distributing P,
we get
= k₃x³ + k₂x² + k₁x + k₀.
Expand the left-hand side into the right-hand side.
The numbers to the right of the letters j and k are subscripts corresponding to the powers of x.
(j₀ and k₀ are constant terms.)