Gujarat BoardEnglish MediumSTD 9MathsPolynomials3 Marks
Question
Factorise:
$3x^3 - x^2 - 3x + 1$
✓
Answer
Let $p(x)=3 x^3-x^2-3 x+1$
Constant term of $p ( x )=-1$
$\therefore$ Factors of $1$ are $\pm 1$
By trial, we find that $p(1)=0$, so $(x-1)$ is a factor of $p(x)$.
Now, we see that $3 x^3-x^2-3 x-1$
$=3 x^2-3 x^2+2 x^2-2 x-x+1$
$=3 x^2(x-1)+2 x(x-1)-1(x-1)$
$=(x-1)\left(3 x^2+2 x-1\right)$
Now, $\left(3 x^2+2 x-1\right)=3 x^2-x^2-3 x-1$ [bi spliting middle term]
$=3 x^2(x+1)-1(x+1)=(x+1)(3 x-1)$
$\therefore 3 x^2-x^2-3 x-1=(x-1)(x+1)(3 x-1)$
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