how do you work this out?

Question

Use grouping to factor the polynomial
tc^2 - tb^2 - qc^2 + qb^2.

Answer 1

(t - q) c² + (q- t) b²
(t - q) c² - (t - q) b²
(t - q) (c² - b²)
(t - q) (c - b) (c + b)

Answer 2

tc^2 - tb^2 - qc^2 + qb^2

(tc^2 - tb^2) + (- qc^2 + qb^2)

factor out GCF
t(c^2 - b^2) + -q(c^2 - b)^2

t(c^2 - b^2) - q(c^2 - b^2)

factor out c^2 - b^2
(c^2 - b^2) (t - q)

c^2 - b^2 is the difference of perfect square. Use this formula:
a^2 - b^2 = (a - b) (a + b)

(c - b) (c + b) (t - q)

Answer 3

I would factor by grouping; i.e., group the first two terms and the last two terms, then factor out the common factor of each two groups:

tc² - tb² - qc² + qb²
(tc² - tb²) - (qc² - qb²)
t(c² - b²) - q(c² - b²)
(t-q)(c - b)(c + b)

Answer 4

just group like terms using the distributive property:
ab + ac = a(b+c)

so:

tc^2 - tb^2 - qc^2 + qb^2
= t ( c^2 - b^2) - q ( c^2 - b^2)

= ( t - q ) * (c^2 - b^2)

And you're done.

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Answer 5

tc^2 - tb^2 - qc^2 + qb^2
(tc^2 - qc^2) - (tb^2 + qb^2)
[c^2(t-q)] - [b^2(t + q)]