BIDMAS

$x(x+1)$ $=x\times x $ $+\, x\times 1$

$\;\;\;=x^2 + x$

$a(b+c)$ $=a\times b $ $+ \,a\times c$

$\;\;\;\;=ab + bc$

Distributive law: $\;a(b+c)=ab + bc$

$x(x+1)$ $=x\times x + x\times 1\qquad $

$\;\;\,\qquad\qquad \qquad =x^2 + x$ $\qquad $Unlike terms👈

$2x(4x+3)$ $=2x\times 4x$ $+\, 2x\times 3\qquad \qquad \qquad$

$\qquad \quad\,=8x^2 + 6x$ $\qquad $Unlike terms👈

$x^2(3y+2x)$ $=x^2\times 3y + x^2\times 2x\qquad \qquad \qquad$

$\quad\qquad \quad=3x^2y + 2x^3$ $\qquad $Unlike terms👈

$5x(x-3)$ $=5x\times x - 5x \times 3\qquad \qquad \qquad$

$\;\;\qquad \quad=5x^2 -15x$ $\qquad $Unlike terms👈

$4x(x^2-3)$ $=4x\times x^2 - 4x\times 3\qquad \qquad \qquad\;\;$

$\qquad \quad=4x^3 - 12x$ $\qquad $Unlike terms👈

$-5x(x-6)$ $=-5x\times x + 5x \times 6\qquad \qquad \qquad\;\;$

$\quad \qquad \;\;=-5x^2 +30x$ $\qquad $Unlike terms👈

Consider now the following

$(a+b)(c+d)$ $= a\times c $ $+\, a\times d $ $+\, b\times c $ $+ \,b\times d\quad \;$

$= a c + a d + b c + bd$

$(x+1)(x+2)$ $= x\times x $ $+ \, x\times 2 $ $+\, 1\times x$ $ +\, 1\times 2\qquad\qquad$

$\qquad \quad\;\;= x^2 + 2x +x + 2$ $\qquad $Like terms👈

$\quad\qquad = x^2 + 3x + 2$ $\qquad $Unlike terms👈

Previously we did expansion:

$4(x+1)$ $=4\times x$ $+\,4\times 1 $

$\quad = 4x+4$

Factorising:

$4x+4$ $=4\times x$ $+\,4\times 1 $

$\quad = 4(x+1)$

$6x+8$ $= 2\times 3 x $ $ +\, 2 \times 4$

$\quad =2(3x+4)$

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$x^2+3x$ $=x$ $\left(\right.$ $x$ $+\,3$ $\left.\right)$

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$4x^2+8x$ $=4x$ $\left(\right.$ $x$ $+\,2$ $\left.\right)$

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$6x-9$ $=3$ $\left(\right.$ $2x$ $-\,3$ $\left.\right)$

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$9x^3-8x^2$ $=x^2$ $\left(\right.$ $9x$ $-\,8$ $\left.\right)$

$a^2 - b^2 $ $=(a+b)(a-b)$

We can check this by expanding:

$(a+b)(a-b)$ $=a\times a $ $-\, a\times b $ $+\, b\times a $ $-\, b\times b $

$\quad =a^2 $ $-\, ab $ $+\, b a $ $-\, b^2 $

$\quad =a^2 $ $-\, ab $ $+\, a b $ $-\, b^2 $

$=a^2-b^2\qquad\;\; $

$x^2-16$ $=x^2-4^2 \qquad \qquad \quad\; $

$=\left(x+4\right)$ $\left(x-4\right)$

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$9x^2-25$ $=(3x)^2-5^2 \qquad \qquad \quad\;\;$

$=\left(3x+5\right)$ $\left(3x-5\right)$

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$2x^2-8$ $=2(x^2-4)$ $=2(x^2-2^2)$

$\;\;\,=2\left(x+2\right)$ $\left(x-2\right)$

$\dfrac{x+2}{x^2+7x +10}$ $=\dfrac{x+2}{(x+2)(x+5)}\qquad $

$=\dfrac{1}{x+5}$

👉Always check answer to see if it can be factorised further!