What is (c-1)(c-2)+(c-3)(c-4) ?

The solution to (c-1)(c-2)+(c-3)(c-4) is 2c^2-10c+14

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simplify (c-1)(c-2)+(c-3)(c-4)

simplify

(c - 1) (c - 2) + (c - 3) (c - 4)

2 c^{2} - 10 c + 14

Solution steps

(c - 1) (c - 2) + (c - 3) (c - 4)

Expand

(c - 1) (c - 2) : c^{2} - 3 c + 2

= c^{2} - 3 c + 2 + (c - 3) (c - 4)

Expand

(c - 3) (c - 4) : c^{2} - 7 c + 12

= c^{2} - 3 c + 2 + c^{2} - 7 c + 12

Group like terms

= c^{2} + c^{2} - 3 c - 7 c + 2 + 12

Add similar elements:

c^{2} + c^{2} = 2 c^{2}

= 2 c^{2} - 3 c - 7 c + 2 + 12

Add similar elements:

- 3 c - 7 c = - 10 c

= 2 c^{2} - 10 c + 2 + 12

Add the numbers:

2 + 12 = 14

= 2 c^{2} - 10 c + 14

z^{2} - \frac{3}{4} z - \frac{1}{8} = 0

\frac{1}{64} x^{3} - \frac{8}{1000} y^{3}

- (\frac{1}{3})^{- 1}

4 < 3 x - 2 \leq 10

p^{2} + 2 p + 1 = 0

What is (c-1)(c-2)+(c-3)(c-4) ?
The solution to (c-1)(c-2)+(c-3)(c-4) is 2c^2-10c+14