For these sorts of factorisations involving rotation of variables, it would be a good idea to rearrange regarding only a specific variable. Example 1 Factorise \( a^2+a^2c+ac^2-b^3-b^2c-bc^2 \). \( \begin{align} &= (ac^2-b^c)+(a^2c-b^2c)+(a^3-b^3) \\ &= (a-b)c^2+(a^2-b^2)c+(a^3-b^3) \\ &= (a-b)c^2+(a-b)(a+b)c+(a-b)(a^2+ab+b^2) \\ &= (a-b)\left[ c^2+(a+b)c+(a^2+ab+b^2) \right] \\ &= (a-b)(a^2+b^2+c^2+ab+bc+ca) \end{align} \) Example 2 Factorise \( x^3+3ax^2+(3a^2-b^2)x+a^3-ab^2 \). \( […]

# Tag Archives: Factorise

# Factorising Quartic Expressions with two Quadratic Factors and a Remainder, such as \( (x^2+3x-2)(x^2+3x+4)-27 \) \( (x^2-8x+12)(x^2-7x+12)-6x^2 \)

Example 1 Factorise \( (x^2+3x-2)(x^2+3x+4)-27 \). \( \require{AMSsymbols} \begin{align} &= \left[ (\bbox[yellow]{x^2+3x})-2 \right] \left[ (\bbox[yellow]{x^2+3x})+4 \right]-27 \\ &= (\bbox[yellow]{x^2+3x})^2 + 2(\bbox[yellow]{x^2+3x})-8-27 \\ &= (\bbox[yellow]{x^2+3x})^2 + 2(\bbox[yellow]{x^2+3x})-35 \\ &= (\bbox[yellow]{x^2+3x}+7)(\bbox[yellow]{x^2+3x}-5) \end{align} \) Example 2 Factorise \( (x^2-8x+12)(x^2-7x+12)-6x^2 \). \( \require{AMSsymbols} \begin{align} &= \left[ (x^2+12)-8x \right] \left[ (x^2+12)-7x \right]-6x^2 \\ &= (x^2+12)^2 -15x(x^2+12) + 56x^2-6x^2 \\ &= […]

# Factorising Quartics with Four Factors and a Remainder such as \( (x-1)(x-3)(x+2)(x+4)+24 \)

Example 1 Factorise \( (x-1)(x-3)(x+2)(x+4)+24 \). \( \require{AMSsymbols} \begin{align} &= (x-1)(x+2) \times (x-3)(x+4) + 24 \\ &= (x^2+x-2) \times (x^2+x-12) +24 \\ &= \left[(x^2+x)-2\right] \times \left[(x^2+x)-12\right] +24 \\ &= (x^2+x)^2-14(x^2+x)+24+24 \\ &= (x^2+x)^2\bbox[aqua]{-14(x^2+x)}+48 \end{align} \) \( \require{AMSsymbols} \begin{array} {ccr} &\bbox[yellow]{x^2+x} &\bbox[pink,3px]{-6} &\bbox[pink]{-6(x^2+x)} \\ &\bbox[pink]{x^2+x} &\bbox[yellow,3px]{-8} &\bbox[yellow]{-8(x^2+x)} \\ \hline &&&\bbox[aqua]{-14(x^2+x)} \end{array} \) \( \require{AMSsymbols} \begin{align} &= […]

# Factorise using Double Common Factors

Transcript What I’m gonna do whenever you have four terms, I’m gonna try to group them into twos okay? So I just put like in just think those brackets are invisible. I’m just going to put brackets in there, so I’m going to do factorize this by individually and this individually. So I’m going to […]

# Factorise Common Factors in Brackets and Indices

Transcript Factorize a plus b squared minus 2a plus b. All right, so you can see that a plus b is common. Because this is one term and this is another. So a plus b is common in those two terms. Now what I’m going to do is separate a plus b squared because some […]

# Factorise Common Factors in Brackets

Transcript This term and this term what is common? What is the common factor? Can you guys see that a plus b is it in both of them? So a plus b is common. So we take the a plus b out the front like that. So we took a plus b, so we just […]

# Factorise using Common Factors and Indices

Transcript Again guys what I’m going to do is trying to expand that out in an expansion in expanded form. So I’ve got one we’ve got two b’s because it’s b squared so I put bb and three c, so ccc, because it’s c cubed right? I just put it like that you don’t have […]

# Factorise using Common Factors

Transcript Guys in these questions you have to ask yourself what is common? Or what is the common factor? Okay? Factor is the keyword here when you’re factorizing that’s why it’s factorizing okay? So in 2a and ab what is the common factor? What is common? So what’s commonly ask yourself what’s common.See how they […]

# Algebraic Factorisation with Exponents (Indices)

$\textit{Factorisation}$ We first look for $\textit{common factors}$ and then for other forms such as $\textit{perfect squares}$, $\textit{difference of two squares}$, etc. Example 1 Factorise $2^{n+4} + 2^{n+1}$. \( \begin{align} \displaystyle&= 2^{n+1} \times 2^{3} + 2^{n+1} \\&= 2^{n+1}(2^{3} + 1) \\&= 2^{n+1} \times 9\end{align} \) Example 2 Factorise $2^{n+3} + 16$. \( \begin{align} \displaystyle&= 2^{n+3} + […]

# Factorising Harder Quadratics

Algebra – Harder Factorise or Factoring can be done by collecting common factors and using sums and products of factors.$$ \large x^2-(a+b)x + ab = (x-a)(x-b)$$ Question 1 Factorise \( (x^2-3x)^2-2x^2 + 6x-8 \). \( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}&= (x^2-3x)^2-2(x^2-3x)-8 \\&= A^2-2A-8 &\color{red} \text{let } A = x^2-3x \\&= (A + 2)(A-4) \\&= (x^2-3x + […]