Question

Obtain all other zeroes of the polynomial x4 +5x3 -6x2 – 32 x +32, if two of its zeroes are 1 and -4..

Answer 1

EXPLANATION.

Zeroes of the polynomial.

⇒ x⁴ + 5x³ - 6x² - 32x + 32.

If two of its zeroes are 1  and  - 4.

As we know that,

Zeroes of the equation.

⇒ x = 1.

⇒ x - 1 = 0. - - - - - (1).

⇒ x = - 4.

⇒ x + 4 = 0. - - - - - (2).

Products of the zeroes of the quadratic equation.

⇒ (x - 1)(x + 4).

⇒ x² + 4x - x - 4.

⇒ x² + 3x - 4.

Divide :

⇒ x⁴ + 5x³ - 6x² - 32x + 32.  by   x² + 3x - 4.

We get,

⇒ x² + 2x - 8.

Now, factorizes the equation into middle term splits, we get.

⇒ x² + 2x - 8 = 0.

⇒ x² + 4x - 2x - 8 = 0.

⇒ x(x + 4) - 2(x + 4) = 0.

⇒ (x - 2)(x + 4) = 0.

⇒ x = 2  and  x = - 4.

All zeroes of the polynomial = 1, -4, 2, -4.

Answer 2

Given :-

$\sf x^{4} + 5x^3-6x^{2} -32x+32$

To Find :-

Other two zeroes

Solution :-

Since highest power is 4. So, It will have two more zeroes

x = 1

x - 1 = 0

x = -4

x - (-4) = 0

x + 4 = 0

Now

Multiply both

(x - 1)(x + 4)

(x × x) + (4 × x) - (1 × x) - (1 × 4)

x² + 4x - x - 4

x² + 3x - 4

Divide x⁴ + 5x³ - 6x² - 32x + 32/x² + 3x - 4

=  x² + 2x - 8

Now

x² + (4x - 2x) - 8

x² + 4x - 2x - 8

x(x + 4) - 2(x + 4)

(x - 2)(x + 4)

Either

x - 2 = 0

x = 0 + 2

x = 2

or

x + 4 = 0

x = 0 - 4

x = -4

[tex][/tex]