Question

factorise
$14 {x}^{4} - 37 {x}^{3} - 72 {x}^{2} - 17x + 4$
​

Answer 1

Answer:

I WILL JUST GIVE A HINT... PLEASE SOLVE IT ON YOUR OWN...

$- 1$

IS A ROOT OF THE EQUATION .. SO

DO LONG DIVISION WITH

$x + 1$

THEN U WILL GET A CUBIC EQUATION...U WILL BE ABLE TO SOLVE IT...

Answer 2

Answer:

(x + 1)(x + 1/2)(x - 1/7)(x - 4)

Step-by-step explanation:

Let check (1) and (-1) for zeroes:

$14(1)^{4}-37(1)^{3}-72(1)^{2}-17(1)+4$ < 0

$14(-1)^{4}-37(-1)^{3}-72(-1)^{2}-17(-1)+4$ = 0

($14x^{4}-37x^{3}-72x^{2}-17x+4$) ÷ (x + 1) = $14x^{3}-51x^{2}-21x+4$

4 is another zero of polynomial

($14x^{3}-51x^{2}-21x+4$) ÷ (x - 4) = 14x² + 5x - 1 = $(x-\frac{1}{7})(x+\frac{1}{2})$

$14x^{4}-37x^{3}-72x^{2}-17x+4$ = (x + 1)(x + 1/2)(x - 1/7)(x - 4)