Four-term common factors

Group four terms

ax + bx + kay + kby = x(a + b) + ky(a + b) = (a + b)(x + ky). The matching bracket is a factor, just as a numerical common divisor would be.

The grouping must retain every sign

For 2x + 3y + 2kx + 3ky, group x terms and y terms to get 2x(1 + k) + 3y(1 + k) = (1 + k)(2x + 3y). You can also group the first pair and last pair to get the same factorisation. Compare complete brackets rather than cancelling individual added terms.

Check your understanding 1 A matching bracket is one factor

Factor ax + ay-bx-by.

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Group as a(x + y) − b(x + y), giving (a − b)(x + y). Expanding restores ax + ay − bx − by. The minus sign multiplies both terms in the second bracket.

Extract an overall common factor when useful

A factorisation can be checked independently by multiplication. Matching a few substituted values may expose a mistake, but expanding the complete factors establishes the identity over the original domain.