Factorize-24 x2-41 x-12

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

By Grouping

Factorize:24 ...

Question

Factorize:
24x² − 41x + 12

Open in App

Solution

We have:
$24 x^{2} - 41 x + 12$
We have to split ( $-$ 41) into two numbers such that their sum is ( $-$ 41) and their product is 288, i.e., $24 \times 12$ .
Clearly, $(- 32) + (- 9) = - 41 and (- 32) \times (- 9) = 288$ .

$∴ 24 x^{2} - 41 x + 12 = 24 x^{2} - 32 x - 9 x + 12 = 8 x (3 x - 4) - 3 (3 x - 4) = (3 x - 4) (8 x - 3)$

Suggest Corrections

Similar questions

Q. Find the value of a and b.

\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} \sqrt{5} b

Hence factorise the polynomial :

24 x^{2} - (a + 41) x + (b + 11)

Q. What is the complete factor of

12 x^{3} + 24 x^{2} - 36 x

Q. Write polynomial

P

as a product of linear factors:

P (x) = (x + 2) (20 x^{2} + 41 x + 20)

If $x^{2}$ - 6x + 11 is a factor of bi-quadratic equation $x^{4}$ + $x^{3}$ - 25 $x^{2}$ + 41x + 66 = 0. Find the product of other factor of bi-quadratic equation and x+1.

Q. If

\frac{x^{4} + 24 x^{2} + 28}{(x^{2} + 1)^{3}}

= \frac{A x + B}{x^{2} + 1} + \frac{C x + D}{(x^{2} + 1)^{2}} + \frac{E x + F}{(x^{2} + 1)^{3}}

then

A =

Join BYJU'S Learning Program

Factorize: 24x2 − 41x + 12

We have: 24x2-41x+12 We have to split (-41) into two numbers such that their sum is (-41) and their product is 288, i.e., 24×12. Clearly, -32+-9=-41 and -32×-9=288. ∴24x2-41x+12=24x2-32x-9x+12 =8x3x-4-33x-4 =3x-48x-3

Factorize:
24x² − 41x + 12