Question

Divide:-
44(x^4 - 5x^3 -24x^2) by 11x(x-8)​

Answer 1

Answer:

Let

f(x)=x4−5x3+11x2−13x+6

We get that

f(1)=1−5+11−13+6=0

and

f(2)=16−5×8+11×4−13×2+6=16−40+44−26+6=0

Hence, we have that

f(x)=(x−1)(x−2)(x2+ax+b)

Plugging in x=0, we get that

f(0)=2b=6⟹b=3.

Plugging in x=3, we get that

f(3)=2(9+3a+3)=81−5×27+11×9−13×3+6

This gives us that

6a+24=12⟹a=−2

Hence, we have that

f(x)=(x−1)(x−2)(x2−2x+3)

Hence,

f(x)=0⟹x=1 or x=2 or x2−2x+3=0

x2−2x+3=0⟹(x−1)2+2=0⟹x=1±i2–√

Hence, the roots ar

Answer 2

Question:-

$44( {x}^{4} - 5 {x}^{3} - 24 {x}^{2}) \div 11x(x - 8)$

Solution:-

Factorising

$44({x}^{4} - {5x}^{3} - 24 {x}^{2}) \: we \: get$

$44( {x}^{4} - 5 {x}^{3} - 24 {x}^{2}) = 2 \times 2 \times 11 \times {x}^{2}( {x}^{2} - 5x - 24)$

(Taking the common factor x^2 out of the bracket)

$= 2 \times 2 \times 11 \times {x}^{2} ( {x}^{2} - 8x + 3x - 24)$

$= 2 \times 2 \times 11 \times {x}^{2} (x(x - 8) + 3(x - 8)$

$= 2 \times 2 \times 11 \times {x}^{2} (x + 3)(x - 8)$

Therefore,

$44( {x}^{4} - {5x}^{3} - 24 {x}^{2}) \div 11x(x - 8)$

$= \frac{2 \times 2 \times 11 \times x \times x \times (x + 3) \times (x - 8)}{11 \times x \times (x - 8)}$

$= 2 \times 2 \times x(x + 3) = 4x(x + 3)$

___________________________________

We cancel the factors 11, x & ( x-8 ) common to both the numerator & denominator.

___________________________________