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 Question from bailey, a student: Factorise f(a)=(6a^2-5a+1)(8a^2-6a+1)(12a^2-7a+1) thus find [f(a)]^1/2

Hi Bailey,

I can help get you started. Look at the first factor

$6a^2 - 5a + 1.$

$6$ can be factored as $6 \times 1$ or $3 \times 2$ and hence, if $6a^2 - 5a + 1$ can be factored it is

$6a^2 - 5a + 1 = (6a \pm \mbox{?})(a \pm \mbox{?})$

or

$6a^2 - 5a + 1 = (3a \pm \mbox{?})(2a \pm \mbox{?}).$

$1$ factors as $1 \times 1$ and hence the question marks are all $1.$ Since the constant term is positive and the middle term $(-5a)$ is negative the signs must be negative and hence if $6a^2 - 5a + 1$ can be factored it is

$6a^2 - 5a + 1 = (6a - 1)(a -1)$

or

$6a^2 - 5a + 1 = (3a -1)(2a -1).$

Which one gives a middle term of $-5a?$

Now you factor $(8a^2-6a+1)$ and $(12a^2-7a+1).$

Penny

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