Find the HCF of $6 x^{3} + 13 x^{2} + 6 x, 8 x^{3} + 6 x^{2} - 9 x$ and $2 x^{3} + 3 x^{2}$ .

Open in App

Solution

Observe that
$6 x^{3} + 13 x^{2} + 6 x = x (6 x^{2} + 13 x + 6) = x (2 x + 3) (3 x + 2)$
$8 x^{3} + 6 x^{2} - 9 x = x (8 x^{2} + 6 x - 9) = x (2 x + 3) (4 x - 3)$ ,
$2 x^{3} + 3 x^{2} = x^{2} (2 x + 3)$ .
Here we can find the HCF of any two polynomials and then use it to find the HCF of all the three, For example, $x (2 x + 3)$ is the HCF of $6 x^{3} + 13 x^{2} + 6 x$ and $2 x^{3} + 3 x^{2}$ . Now HCF of $x (2 x + 3)$ and $8 x^{3} + 6 x^{2} - 9 x$ is simply $x (2 x + 3)$ . Therefore the HCF of three polynomials is $x (2 x + 3)$

Suggest Corrections

Similar questions

f (x) = \frac{3}{10} x^{4} - \frac{4}{5} x^{3} - 3 x^{2} + \frac{36}{5} x + 11

f (x) = \frac{x^{4}}{4} + \frac{2}{3} x^{3} - \frac{5}{2} x^{2} - 6 x + 7

f (x) = {\{x (x - 2)\}}^{2}

p (x) = x^{4} - 2 x^{3} - 15 x^{2}

q (x) = x^{3} - 9 x

{lim}_{x \to 2} \frac{x^{3} - 6 x^{2} + 11 x - 6}{x^{2} - 6 x + 8}

{lim}_{x \to 2} \frac{x^{3} - 6 x^{2} + 11 x - 6}{x^{2} - 6 x + 8}

\frac{6 x^{2} + 9 x}{3 x^{2} - 12 x}

Join BYJU'S Learning Program