What is expand (c)(-jf-jfa-j-ej-j^2c-ja+ejb) ?

The solution to expand (c)(-jf-jfa-j-ej-j^2c-ja+ejb) is -jcf-jcfa-jc-ejc-j^2c^2-jca+ejcb

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expand (c)(-jf-jfa-j-ej-j^2c-ja+ejb)

expand

(c) (- j f - j f a - j - e j - j^{2} c - ja + e jb)

- j c f - j c f a - j c - e j c - j^{2} c^{2} - j c a + e j c b

Solution steps

(c) (- j f - j f a - j - e j - j^{2} c - ja + e jb)

Distribute parentheses

= (c) (- j f) + (c) (- j f a) + (c) (- j) + (c) (- e j) + (c) (- j^{2} c) + (c) (- ja) + (c) e jb

Apply minus-plus rules

+ (- a) = - a

= - j (c) f - j (c) f a - j (c) - e j (c) - j^{2} (c) c - j (c) a + e j (c) b

Simplify

- j (c) f - j (c) f a - j (c) - e j (c) - j^{2} (c) c - j (c) a + e j (c) b : - j c f - j c f a - j c - e j c - j^{2} c^{2} - j c a + e j c b

= - j c f - j c f a - j c - e j c - j^{2} c^{2} - j c a + e j c b

\frac{7 \cdot 2 ^{(k + 1)} + ( - 1 ) ^{(k + 2)}}{3}

(- 3, 2, 2) (1 - t) + (1, - 2, 5) t

(4 + 2 i)^{\circ} (1 + 2 i)

(x + 1) d y

\frac{ln ( x + a ) ( x + a ) + x}{x + a}

What is expand (c)(-jf-jfa-j-ej-j^2c-ja+ejb) ?
The solution to expand (c)(-jf-jfa-j-ej-j^2c-ja+ejb) is -jcf-jcfa-jc-ejc-j^2c^2-jca+ejcb