Question
Factorise: $10x^2– 18x^3+ 14x^4$
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Answer
$10x^2= 2 \times 5 \times x \times x$
$ 18x^3= 2 \times 3 \times 3 \times x \times x \times x$
$ 14x^4= 2 \times 7 \times x \times x \times x \times x$
The common factors of the three terms are $2, x$ and $x.$
Therefore, $10x^2– 18x^3+ 14x^4= (2 \times x \times x \times 5) – (2 \times x \times x \times 3 \times 3 \times x) + (2 \times x \times x \times 7 \times x \times x)$
$= 2 \times x \times x \times [5 – (3 \times 3 \times x) + (7 \times x \times x)]$ (combining the three terms)
$= 2x^2(5 – 9x + 7x^2) = 2x^2(7x^2- 9x + 5)$
$ 18x^3= 2 \times 3 \times 3 \times x \times x \times x$
$ 14x^4= 2 \times 7 \times x \times x \times x \times x$
The common factors of the three terms are $2, x$ and $x.$
Therefore, $10x^2– 18x^3+ 14x^4= (2 \times x \times x \times 5) – (2 \times x \times x \times 3 \times 3 \times x) + (2 \times x \times x \times 7 \times x \times x)$
$= 2 \times x \times x \times [5 – (3 \times 3 \times x) + (7 \times x \times x)]$ (combining the three terms)
$= 2x^2(5 – 9x + 7x^2) = 2x^2(7x^2- 9x + 5)$
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