Question

How do you factor `x^2-4x-21`?

Answer 1

`(x-7)(x+3)`

Explanation:

`"one way is to 'split' the middle term "`

`"that is " -4x=-7x+3x`

`rArrx^2-7x+3x-21`

`"factorise each 'pair' of terms"`

`=color(red)(x)(x-7)color(red)(+3)(x-7)`

`"take out the common factor " (x-7)`

`=(x-7)(color(red)(x+3))`

`rArrx^2-4x-21=(x-7)(x+3)`

`color(blue)"Another way is"`

`"the expression is in the form " ax^2+bx+c`

`"to factor consider the factors of the product ac"`
`"which also sum to give b"`

`"here " a=1,b=-4" and "c=-21`

`ac=1xx-21=-21`

`"the factors of - 21 which sum to - 4 are " -7" and "3`

`rArrx^2-4x-21=(x-7)(x+3)`

Answer 2

Because the quadratic

`x^2-4x-21`

has `1` as its coefficient of `x^2` we will start with two brackets like this:

`(x+-a) (x+-b)`

what is left to be determined are the values for `a " & " b`

to decide what these numbers are we look for two numbers that multiply to give the constant `ie. -21` but add up to the coefficient of `x` `ie. -4`

this we do by inspection, taking note of teh signs

`a=1, b=-21 " sum "=1-21=20 " incorrect"`

`a=-3, b=7 " sum " =4" correct value but incorrect sign"`

`a=3, b=-7" sum "=-4 " correct value and sign"`

`:.x^2-4x-21=(x-7)(x+3)`