Question

Factorise fully

a) \(6 x y-2 x\)
b) \(x^2(y+3)-(y+3)\)
c) \(4 x^2-y^2\)
d) \(x^2-11 x+18\)

Answer 1

a) We can factor out a common factor of 2x from both terms:
$$6xy - 2x = 2x(3y-1)$$
b) We can factor out a common factor of (y+3) from both terms:
$$x^2(y+3) - (y+3) = (y+3)(x^2-1)$$
Now we can factor the expression inside the second set of parentheses using the difference of squares formula:
$$(y+3)(x^2-1) = (y+3)(x-1)(x+1)$$
c) This expression is already in the form of a difference of squares, so we can apply the formula:
$$4x^2 - y^2 = (2x+y)(2x-y)$$
d) We can factor this expression by finding two numbers that multiply to 18 and add to -11. These numbers are -2 and -9, so we can write:
$$x^2 - 11x + 18 = (x-2)(x-9)$$