Question

factorise a² – 21a + 108​

Answer 1

Answer:

$a^2-21a+108\\= \left(a^2-9a\right)+\left(-12a+108\right)\\= a\left(a-9\right)-12\left(a-9\right)v\\= \left(a-9\right)\left(a-12\right)$

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Answer 2

$\bf \huge {\underline {\underline \red{AnSwEr}}}$

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$\bf \pink{ {a}^{2} - 21a + 108}$

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Write 21a as a difference

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$\bf \purple{ {a}^{2} - 9a - 12a + 108}$

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Factor out a and -12 from expression

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$\bf \pink{a(a - 9) - 12(a - 9)}$

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Factor out a - 9 from expression

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$\bf \purple{(a - 9)(a - 12)}$

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Additionally

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How to factorize :-

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• $\bf\green {Step 1:}$Find two numbers that multiply to give ac (in other words a times c), and add to give b.

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• $\bf\green {Step 2:}$ Rewrite the middle with those numbers

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• $\bf\green {Step 3:}$ Factor the first two and last two terms separately:

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• $\bf\green {Step 4:}$ If we've done this correctly, our two new terms should have a clearly visible common factor.