GCSE(H),

Factorising the expression 9*x*^{2} - 25*x* - 6 includes a coefficient for the *x*^{2} term (9). The factorised expression will therefore be of the form (*px* + *q*)(*rx* + *s)*, where *p*, *q*, *r* and *s* are integers.

There are three known facts about the factors:

*p* x *r* = 9; therefore *p* and *r* are factors of 9;

*q* x *s* = -6; therefore *q* and *s* are factors of -6;

*ps* + *qr* = -25.

The factors of 9 are (1, 9) and (3, 3). The factors of -6 are (1, -6), (2, -3), (3, -2), (6, -1), (-1, 6), (-2, 3), (-3, 2), (-6, 1). The answer involves a combination of these factor sets: often they can be determined by **inspection**; that is, looking for an immediate answer.

The expression factorises to (9*x* + 2)(*x* - 3).

1. Factorise 8*x*^{2} + 22*x* + 5.

Answer: (4*x* + 1)(2*x* + 5)

The factors of 5 are (1, 5). The factors of 8 are (1, 8), (2, 4), (4, 2). Set out a table to determine which combinations of these factors give 22.

The coefficients of *x* are 2 and 2; the integers are 2 and 3.

2. Factorise -21*x*^{2} + 29*x* - 10.

Answer: (-7*x* + 5)(3*x* - 2)

Look at the positive factors only for 21: (1, 21), (3, 7), and 10: (1, 10), (2, 5), to see which combination gives an answer of 29:

(1, 10) | (2, 5) | |

(1, 21) | 1x21+10x1=31 | 2x21+5x1=47 |

(3, 7) | 1x7+10x3=37 | 2x7+5x3=29 |

The combination is based on (7*x* + 5)(3*x* + 2); check positive/negative combinations to obtain (-7*x* + 5)(3*x* - 2)

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