# Factorise these mixed expressions using group factorising method. a) (x+4)^2 +5(x+4) b) (y+1)^2 + 4(y+1)

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Factorise in algebra terms means multiply out brackets. Mixed expression in algebra terms refers to an expression that involves terms in the variable of interest with different powersAn expression is just a collection of added and multiplied numbers and letters and a variable is represented by a letter, like x. They are represented by a letter to indicate that they can take the value of any number. For example, x could be the number of apples a person buys each week.

a) Factorise the mixed expression

`(x+4)^2 + 5(x+4)`

It is a mixed expression, because there will be (once we've factorised) `x^2` terms (squared terms, where we raise `x` to the power 2) and plain `x` terms (linear terms, where we raise `x` to the power 1). Also, there will be constant terms, where x is raised to the power 0 (anything raised to the power zero gives 1).

To factorise, note that `(x+4)^2 = (x+4)(x+4)` because we are multiplying the value `(x+4)` by itself (raising it to the power 2, or squaring it). Also, to see what is happening, we can put a bracket round the number 5 in the second part of the expression. So, we have that the expression is given by

`(x+4)(x+4) + (5)(x+4)`

Now multiply out the brackets, making sure each term in each bracket is multiplied once by each term in the neighboring bracket:

`(x^2 + 4x + 4x + 4^2) + (5x + 4) `

Now gather terms and simplify within the two parts of the expression (which I've bracketed to indicate where the two parts separate). The second part doesn't need anything doing to it, so leave that as it is:

`(x^2 + 8x + 16) + (5x + 4)`

Now remove the brackets from the separated parts:

`x^2 + 8x + 16 + 5x + 4`

and finally simplify by collecting like terms (terms that share the same power of x - squared terms of power 2, then linear terms of power 1, then constants of power 0):

`x^2 + 13x + 20`   answer

b) Work this through similarly to a), though here the variable is y and not x. Though the variable in this expression is labelled a different letter, treat it in exactly the same way as x - just an unknown, that can potentially take the value of any number.

`(y+1)^2 + 4(y+1)`

Write out fully, converting power terms to brackets multiplied together:

`(y+1)(y+1) + (4)(y+1)`

Multiply out brackets:

`(y^2 + y + y + 1^2) + (4y + 4)`

Simplify within the two parts of the expression:

`(y^2 + 2y + 1) + (4y + 4)`

Combine the two parts by removing brackets:

`y^2 + 2y + 1 + 4y + 4`

Simplify into terms with different powers in y - squared, linear, constant:

`y^2 + 6y + 5`  answer