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\$\\simplify{x^2+{linear}x+{const}}\$ = [[0]].

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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Since \$(x+a)(x+b)=x^2+(a+b)x+ab\$, when we are factorising a quadratic, such as \$x^2+cx+d\$, we must find the numbers \$a\$ and \$b\$ such that \$c=a+b\$ and \$d=ab\$.

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In the case of \$\\simplify{x^2+{linear}x+{const}}\$ we ask

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what two numbers add to give \$\\var{linear}\$ and multiply to give \$\\var{const}\$?

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Therefore the numbers must be \$\\var{a}\$ and \$\\var{b}\$, that is

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\$\\simplify{x^2+{linear}x+{const}}=(\\simplify{x+{a}})(\\simplify{x+{b}}).\$

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You can check this by expanding the binomial product.

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Factorise the following into linear factors. That is, write the quadratic as a product of terms that look like \$ax+b\$ where \$a\$ and \$b\$ are real numbers.

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