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Follow the steps outlined above to help you factorise the questions.

Click on 'Try another question like this' to try other questions using different numbers.

Alternatively, click on this link to watch a video on factorising expressions.

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$\\var{c[0]}-\\var{c[1]}x^2=$

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Først finner vi største felles faktor, i dette tilfelle $2$. Vi setter den så utenfor parentes: $2(\\;\\;\\;\\;)$

Deretter må vi finne ut hva som står igjen i leddene inne i parentesen.

For å finne dette kan vi dele hvert ledd med den felles faktoren:

$2\\big(\\frac{\\var{c[0]}}{2}-\\frac{\\var{c[1]}}{2}x^2\\big)=2\\big(\\simplify{{c[0]}/2}-\\simplify{{c[1]}/2}x^2\\big)$

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Your answer must be factorised

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$\\var{c[2]}ab+\\var{c[3]}bc=$

Først finner vi største felles faktor, i dette tilfelle $2b$. Vi setter den så utenfor parentes: $2b(\\;\\;\\;\\;)$

Deretter må vi finne ut hva som står igjen i leddene inne i parentesen.

For å finne dette kan vi dele hvert ledd med den felles faktoren:

$2b\\big(\\frac{\\var{c[2]}ab}{2b}+\\frac{\\var{c[3]}bc}{2b}\\big)=2b\\big(\\simplify{({c[2]}/2)*a}+\\simplify{({c[3]}/2)*c}\\big)$

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Your answer must be factorised

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$\\var{c[4]}a^2+\\var{c[5]}ab=$

Først finner vi største felles faktor, i dette tilfelle $2a$. Vi setter den så utenfor parentes: $2a(\\;\\;\\;\\;)$

Deretter må vi finne ut hva som står igjen i leddene inne i parentesen.

For å finne dette kan vi dele hvert ledd med den felles faktoren:

$2a\\big(\\frac{\\var{c[4]}a^2}{2a}+\\frac{\\var{c[5]}ab}{2a}\\big)=2a\\big(\\simplify{({c[4]}/2)*a}+\\simplify{({c[5]}/2)*b}\\big)$

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Your answer must be factorised

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$pq^3-p^3q=$

Først finner vi største felles faktor, i dette tilfelle $pq$. Vi setter den så utenfor parentes: $pq(\\;\\;\\;\\;)$

Deretter må vi finne ut hva som står igjen i leddene inne i parentesen.

For å finne dette kan vi dele hvert ledd med den felles faktoren:

$pq\\big(\\frac{pq^3}{pq}-\\frac{p^3q}{pq}\\big)=pq(q^2-p^2)$

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Your answer must be factorised

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$\\var{c2[0]}x^2y+\\var{c2[1]}xy^4=$

Først finner vi største felles faktor, i dette tilfelle $3xy$. Vi setter den så utenfor parentes: $3xy(\\;\\;\\;\\;)$

Deretter må vi finne ut hva som står igjen i leddene inne i parentesen.

For å finne dette kan vi dele hvert ledd med den felles faktoren:

$3xy\\big(\\frac{\\var{c2[0]}x^2y}{3xy}+\\frac{\\var{c2[1]}xy^4}{3xy}\\big)=3xy\\big(\\simplify{({c2[0]}/3)*x}+\\simplify{({{c2}[1]}/3)*y^3}\\big)$

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Your answer must be factorised

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$\\var{c3[0]}p^3q-\\var{c3[1]}p^2q^2+\\var{c3[2]}pq^3=$

Først finner vi største felles faktor, i dette tilfelle $2pq$. Vi setter den så utenfor parentes: $2pq(\\;\\;\\;\\;)$

Deretter må vi finne ut hva som står igjen i leddene inne i parentesen.

For å finne dette kan vi dele hvert ledd med den felles faktoren:

$2pq\\big(\\frac{\\var{c3[0]}p^3q}{2pq}-\\frac{\\var{c3[1]}p^2q^2}{2pq}+\\frac{\\var{c3[2]}pq^3}{2pq}\\big)=2pq\\big(\\simplify{({c3[0]}/2)p^2}-\\simplify{({c3[1]}/2)*p q}+\\simplify{({c3[2]}/2)*q^2}\\big)$

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Your answer must be factorised

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$\\var{c2[2]}lm^2-\\var{c2[3]}l^3m^3+\\var{c2[4]}l^2m^4=$

Først finner vi største felles faktor, i dette tilfelle $3lm^2$. Vi setter den så utenfor parentes: $3lm^2(\\;\\;\\;\\;)$

Deretter må vi finne ut hva som står igjen i leddene inne i parentesen.

For å finne dette kan vi dele hvert ledd med den felles faktoren:

$3lm^2\\big(\\frac{{c2[2]}lm^2}{3lm^2}-\\frac{{c2[3]}l^3m^3}{3lm^2}+\\frac{{c2[4]}l^2m^4}{3lm^2}\\big)=3lm^2\\big(\\simplify{({c2[2]}/3)}-\\simplify{({c2[3]}/3)*l^2m}+\\simplify{({c2[4]}/3)*lm^2}\\big)$

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Your answer must be factorised

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$\\var{firstsquare}-\\var{secondsquare}x^2$ =

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Dette kjenner vi som 3.kvadratsetning, eller konjugatsetningen:

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

Vi kan multiplisere parentesene på høyre side for å vise at det stemmer::

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

$ \"a\" $ og $ \"b\" $ kan være byttet ut med mer sammensatte uttrykk, husk da at hver faktor i leddene skal opphøyes i andre:

$a^2x^4-b^2y^6=(ax^2+by^3)(ax^2-by^3)$

Leddene kan også være kombinasjoner av tall og bokstaver:

$9x^2-16y^4=(3x+4y^2)(3x-4y^2)$

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$\\simplify{{qa}*x^2+{qb}*x+{qc}}$ =

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When doing the opposite of factorisation, i.e. expanding brackets, it is sometimes possible to reveal a quadratic expression such as the one in this question from the product of two binomials where the highest power of the variable is $1$. It follows, therefore, that some quadratic expressions can be factorised into the product of two binomials. When this is possible, we say that the quadratic expression can be 'factorised with integers'.

For example, $-6x^2+x+2=(2x+1)(-3x+2)$

In more general terms, the quadratic expression can be shown as:

$ax^2+bx+c$ and the product of binomials could be $(dx+e)(fx+g)$

When expanding the brackets of the binomial products, the quadratic is formed as:

$(d\\times f)x^2+((d\\times g)+(e\\times f))x+(e\\times g)$

Factorising the quadratic with integers, therefore, requires the insight of reverse engineering the products of the coefficients to reflect the pattern shown above.

In this question, the coefficients of the quadratic expression can be analysed as follows:

$a=(d\\times f)=\\var{qa},\\;\\;\\;b=((d\\times g)+(e\\times f))=\\var{qb},\\;\\;\\;c=(e\\times g)=\\var{qc}$

Look for all possible factors of these coefficients and try different variations until you find those which satisfy all conditions.

The result should be:

$d=\\var{q1},\\;\\;\\;e=\\var{q2}\\;\\;\\;f=\\var{q3}\\;\\;\\;g=\\var{q4}$

So that,

$(\\var{q1}\\times\\var{q3})x^2+((\\var{q1}\\times\\var{q4})+(\\var{q2}\\times\\var{q3}))x+(\\var{q2}\\times\\var{q4})$

$\\therefore,\\;\\;\\;\\simplify{{qa}*x^2+{qb}*x+{qc}}=(\\simplify{{q1}x+{q2}})(\\simplify{{q3}*x+{q4}})$

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Faktoriser følgende uttrykk:

Merk: Bruk stjerne (*) til multiplikasjon og hatt (^) for å opphøye i en potens. For eksempel, $xy$ skrives som $x$*$y$, $a(b+c)$ skrives som $a$*$(b+c)$ og $x^2$ som $x$^2.

Under \"Vis trinn\" finnes fullstendig løsning til oppgavene.

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Coefficients in a,b,c (HCF: 2)

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Part f (HCF:2)

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Coefficients in e,f (HCF: 3)

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Factorising polynomials using the highest common factor.

Adapted from 'Factorisation' by Steve Kilgallon.

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