Title: 13. Factorising algebraic expressions
Course: Demonstrating STACK (v4.4.7)

Factorise the following: \[ [value]. \]
Feedback: As can be seen by multiplying out a general, factorised quadratic: \[ (cx+a)(dx+b) = cdx^2+(da+cb)x+ab, \] \(a\) and \(b\) are factors of the constant term, and \(c\) and \(d\) are factors of the \(x^2\) coefficient so there are only a finite number of (integer) cases that are possible. With practice it should be simple enough to find \(a\), \(b\), \(c\) and \(d\). In this example: \[ [value]. \] Equating with the general quadratic above gives: \[{cd = [value]}.\] \[{da+cb = [value]}.\] \[{ab = [value]}.\] The correct values are: \(a = [value]\), and \(b = [value]\), \(c = [value]\) and, \(d = [value]\). By substituting these back into the general quadratic, and factorising constants out, we obtain the factored form: \[[value].\]

Factorise the following: \[ [value] .\]
Feedback: Factorising a quadratic with an \(x^2\) coefficient equal to 1 is a relatively simple task. As can be seen by multiplying out a general, factorised quadratic: \[ (x+a)(x+b) = x^2+(a+b)x+ab \] the task is to find values, \(a\) and \(b\), such that their sum is equal to the \(x\) coefficient and their product is equal to the constant term. In this example: \[ [value]. \] Equating with the general quadratic above gives: \[{a+b = [value]}.\] \[{ab = [value]}.\] With practice you should be able to solve these simultaneous equations very quickly by inspection. We know that \(a\) and \(b\) are factors of [value] so there are just a few cases to check. The correct values are a = [value] and b = [value] (or vice versa). By substituting these back into the general quadratic we obtain the factored form: \[[value].\]

Factorise the following: \[ [value]. \]
Feedback: Factorising a quadratic with an \(x^2\) coefficient equal to 1 is a relatively simple task. As can be seen by multiplying out a general, factorised quadratic: \[ (x+a)(x+b) = x^2+(a+b)x+ab \] the task is to find values, \(a\) and \(b\), such that their sum is equal to the \(x\) coefficient and their product is equal to the constant term. In this example: \[ [value]. \] Equating with the general quadratic above gives: \[{a+b = [value]}.\] \[{ab = [value]}.\] With practice you should be able to solve these simultaneous equations very quickly by inspection. We know that \(a\) and \(b\) are factors of [value] so there are just a few cases to check. The correct values are a = [value] and b = [value] (or vice versa). By substituting these back into the general quadratic we obtain the factored form: \[[value].\]

Factorise the following: \[ [value]. \]
Feedback: Factorising a quadratic with an \(x^2\) coefficient equal to 1 is a relatively simple task. As can be seen by multiplying out a general, factorised quadratic: \[ (x+a)(x+b) = x^2+(a+b)x+ab, \] the task is to find values, \(a\) and \(b\), such that their sum is equal to the \(x\) coefficient and their product is equal to the constant term. In this example: \[ [value]. \] Equating with the general quadratic above gives: \[{a+b = [value]}.\] \[{ab = [value]}.\] With practice you should be able to solve these simultaneous equations very quickly by inspection. We know that \(a\) and \(b\) are factors of [value] so there are just a few cases to check. The correct values are a = [value] and b = [value] (or vice versa). By substituting these back into the general quadratic we obtain the factored form: \[[value].\]

Factorise the following: \[ [value]. \]
Feedback: You can save time in this quadratic by spotting it is of the form, 'square, twice product, square' or \((a+b)^2 = a^2+2ab+b^2 \). Now equating these parts is far simpler: \[ a = [value]. \] \[ b = [value]. \] By substituting this back into \((a+b)^2\) and removing any constant factors we have our factored form: \[[value].\]

Factorise the following: \[ [value] .\]
Feedback: When factorising this expression, it is convenient to notice that it is the difference of two squares. The general rule for the differece of two squares is \(a^2-b^2 = (a-b)(a+b)\). In this example, notice that: \[ a^2 = [value]. \] \[ b^2 = [value]. \] Therefore: \[ a = [value]. \] \[ b = [value]. \] Now, substituting these back into \((a-b)(a+b)\) and removing any constant factors, we obtain: \[ [value] .\]

Factorise the following: \[ [value]. \]
Feedback: Factorising a quadratic with an \(x^2\) coefficient which is not equal to 1 is slightly more complex. As can be seen by multiplying out a general, factorised quadratic: \[ (cx+a)(dx+b) = cdx^2+(da+cb)x+ab. \] \(a\) and \(b\) are still factors of the constant term, and \(c\) and \(d\) are factors of the \(x^2\) coefficient so there are still only a finite number of (integer) cases that are possible. With practice it should still be simple enough to find \(a\), \(b\), \(c\) and \(d\). In this example: \[ [value]. \] Equating with the general quadratic above gives: \[{cd = [value]}.\] \[{da+cb = [value]}.\] \[{ab = [value]}.\] The correct values are: \(a = [value]\), and \(b = [value]\), \(c = [value]\) and, \(d = [value]\). By substituting these back into the general quadratic, and factorising constants out, we obtain the factored form: \[[value].\]

Factorise the following: \[ [value]. \]
Feedback: In this example a clever trick can save you alot of time in factoring this quadratic. Using the fact: \[ [value] = ([value])^2+[value].\] It is now clear that the quadratic is of the form 'square, twice product, square' or \((a+b)^2 = a^2+2ab+b^2 \). Now equating these parts is far simpler: \[ a = [value]. \] \[ b = [value]. \] By substituting this back into \((a+b)^2\) and removing any constant factors we have our factored form: \[[value].\]

Factorise the following: \[ [value]. \]
Feedback: When factorising this expression, it is convenient to notice that it is the difference of two squares. The general rule for the differece of two squares is \(a^2-b^2 = (a-b)(a+b)\). In this example, notice that: \[ a^2 = [value]. \] \[ b^2 = [value]. \] Therefore: \[ a = [value]. \] \[ b = [value]. \] Now, substituting these back into \((a-b)(a+b)\) and removing any constant factors, we obtain: \[ [value] .\]

Factorise the following: \[ [value] .\]
Feedback: Factorising a quadratic with an \(x^2\) coefficient equal to -1 is a relatively simple task. Firstly remove a factor of -1 and proceed as normal. \[ [value] = -([value]). \] As can be seen by multiplying out a general, factorised quadratic: \[ (x+a)(x+b) = x^2+(a+b)x+ab. \] the task is to find values, \(a\) and \(b\), such that their sum is equal to the \(x\) coefficient and their product is equal to the constant term. In this example: \[ -([value]). \] Equating with the general quadratic above gives: \[{a+b = [value]}.\] \[{ab = [value]}.\] With practice you should be able to solve these simultaneous equations very quickly by inspection. We know that \(a\) and \(b\) are factors of [value] so there are just a few cases to check. The correct values are a = [value] and b = [value] (or vice versa). By substituting these back into the general quadratic we obtain the factored form: \[[value].\]

Factorise the following: \[ [value]. \]
Feedback: Firstly notice that: \[ [value] = [value]+[value]+([value])^2 .\] The equation is of the form 'square, twice product, square': \[ a^2+2ab+b^2 = (a+b)^2. \] Where: \[ a^2 = x. \] \[ b^2 = 4y. \] Therefore \((a+b)^2 = (x+4y)^2. \)

Factorise the following: \[ [value]. \]
Feedback: You can factorise the cubic \([value]\) by removing the common factor \([value]\). This leaves: \[ [value]\left([value]\right).\] Now you can simply factorise the quadratic to obtain the fully factored form: \[ [value]\left([value]\right)\left([value]\right). \]

Factorise the following: \[ [value]. \]
Feedback: You can remove a common factor \(2a\) which leaves: \[ 2a\left(25a^2-9b^2\right). \] This is clearly a difference of two squares so the fully factored form is: \[2a\left(5a+3b\right)\left(5a-3b\right).\]

Factorise the following: \[ [value]. \]
Feedback: The trick to factorising this expression is to carry out a substitution, \(u=xy\). Then we obtain the quadratic: \[ u^2-2u+1, \] which is easily solved by normal methods: \[ u^2-2u+1 = \left(u-1\right)^2 = \left(xy-1\right)^2. \]

Factorise the following: \[ [value]. \]
Feedback: Although it would be possible to multiply out the brackets and then factorise, it is far simpler, and far less time consuming to peform two clever substitutions. \[ u = a+b. \] \[ v = a+c.\] This gives: \[ u^2-v^2, \] which is now clearly a difference of two squares. So factorising gives: \[ u^2-v^2 = \left( u+v \right) \left( u-v \right) = \left( a+b+a+c \right) \left( a+b-a-c \right) = \left( 2a+b+c \right) \left( b-c \right). \]

Factorise the following: \[ [value] .\]
Feedback: This quartic equation is really a quadratic in disguise. This becomes clear by using the substitution, \(u=[value]\). The expression becomes: \[[value],\] which can be factorised in the convential way to obtain: \[ [value]. \] Be sure to then substitute \(u=[value]\) back to give your factored form in terms of \([value]\): \[ [value]. \]

Factorise the following:
\[ [value] .\]
Feedback: When factorising this expression, it is convenient to notice that it is the difference of two squares. The general rule for the differece of two squares is \(a^2-b^2 = (a-b)(a+b)\). In this example, notice that: \[ a^2 = [value]. \] \[ b^2 = [value]. \] Therefore: \[ a = [value]. \] \[ b = [value]. \] Now, substituting these back into \((a-b)(a+b)\) and removing any constant factors, we obtain: \[ [value] .\]

Factorise the following: \[ [value]. \]
Feedback: When factorising this expression, simply focus on grouping one letter at a time and you will arrive at the factored form. Firstly, factored out \(a\)'s to give: \[ [value] = [value]. \] Then factorise out \(b\)'s to give: \[ [value]. \] Then factor out \(x+y\) to give: \[ [value]. \]

Factorise the following: \[ [value]. \]
Feedback: When factorising the sum of two cubes it is probbaly easiest to remember the formula: \[ x^3+y^3 = \left(x+y\right)\left(x^2-xy+y^2\right). \] So in this example: \[ [value] = [value] = [value]. \]

Factorise the following: \[ [value]. \]
Feedback: When factorising the difference of two cubes it is probbaly easiest to remember the formula: \[ x^3-y^3 = \left(x-y\right)\left(x^2+xy+y^2\right). \] So in this example, using the formula and factoring out any constants gives: \[ [value] = [value]. \]