Answer :

For finding zeroes,we’ll use the hit and trial method.

For that we’ll put a value which will make our polynomial zero.

Since, here the whole equation is combination of negative positive that means, zero can be achieved by putting the positive or negative values.

So, first we’ll put x = – 1.

⇒ (– 1)^{4} – 2(– 1)^{3} – 7(– 1)^{2} + 8(– 1) + 12 = 1 + 2 – 7 – 8 + 12

= 0

⇒ x = – 1 is one of the zeroes.

Now, we’ll write the equation in terms of (x + 1) to make equation quadratic and hence to find the other zeroes.

⇒ x^{4} – 2x^{3} – 7x^{2} + 8x + 12 = x^{3} (x + 1) – 3x^{2} (x + 1) – 4x(x + 1) + 12(x + 1)

= (x + 1)(x^{3} – 3x^{2} – 4x + 12)

Again, we’ll use hit and trial for the cubic equation.

So, now we’ll put x = 2 in x^{3} – 3x^{2} – 4x + 12

⇒ 2^{3} – 3(2)^{2} – 4(2) + 12 = 8 – 12 + 8 + 12

= 0

⇒ x = 2 is another zero. Now, we’ll write the cubic equation in terms of x – 2.

⇒ x^{4} – 2x^{3} – 7x^{2} + 8x + 12 = (x + 1){ x^{2} (x – 2) – x(x – 2) – 6(x – 2)}

= (x + 1)(x – 2)(x^{2} – x – 6)

Now, using factorization by splitting the middle term method

= (x + 1)(x – 2)(x^{2} – (3 – 2)x – 6)

= (x + 1)(x – 2)(x^{2} – 3x + 2x – 6)

= (x + 1)(x – 2)(x(x – 3) + 2(x – 3))

= (x + 1)(x – 2)(x + 2)(x – 3)

Now, to get the zeroes we’ll put these equals to zero.

⇒ x + 1 = 0, x – 2 = 0, x + 2 = 0, x – 3 = 0

⇒ x = – 1, x = 2, x = – 2 and x = 3 are the zeroes of x^{4} – 2x^{3} – 7x^{2} + 8x + 12

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