Question

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

Answer:

(1)4+5(1)3-6(1)2-32(1)+32

1+5-6-32+32

0

(-4)4+5(-4)3-6(-4)2-32(-4)+32

=0

Answer 2

Given :-

x⁴ + 5x³ - 6x² - 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