# Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficient in each case :

Question : Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficient in each case :

(i)  $$2x^3 + x^2 – 5x + 2$$;  $$1\over 2$$, 1, -2

(ii)  $$x^3 – 4x^2 + 5x – 2$$;  2, 1, 1

Solution :

(i)  Comparing the given polynomials with $$ax^3 + bx^2 + cx + d$$, we get

a = 2, b = 1, c = -5 and d = 2

p($$1\over 2$$) = 2$$({1\over 2})^3$$ + $$({1\over 2})^2$$ – 5($$1\over 2$$) + 2 = $$1\over 4$$ + $$1\over 4$$ – $$5\over 2$$ + 2 = $$0\over 4$$ = 0

p(1) = 2$$(1)^3$$ + $$(1)^2$$ – 5(1) + 2 = 2 + 1 – 5 + 2 = 0

p(-2) = 2$$(-2)^3$$ + $$(-2)^2 – 5(-2) + 2 = -16 + 16 = 0 \(\therefore$$   $$1\over 2$$, 1 and -2 are the zeroes of  $$2x^3 + x^2 – 5x + 2$$.

So,  $$\alpha$$ = $$1\over 2$$, $$\beta$$ = 1 and $$\gamma$$ = -2.

Therefore, $$\alpha$$ + $$\beta$$ + $$\gamma$$ = $$1\over 2$$ + 1 + (-2) = $$-1\over 2$$ = $$-b\over a$$

$$\alpha$$$$\beta$$ + $$\beta$$$$\gamma$$ + $$\gamma$$$$\alpha$$  = ($$1\over 2$$)(1) + (1)(-2) + (-2)($$1\over 2$$) = $$-5\over 2$$ = $$c\over a$$

and  $$\alpha$$$$\beta$$$$\gamma$$ = $$1\over 2$$ $$\times$$ 1 $$\times$$ (-2) = $$-2\over 2$$ = $$-d\over a$$

(ii)  Comparing the given polynomials with $$ax^3 + bx^2 + cx + d$$, we get

a = 1, b = -4, c = 5 and d = -2

p(2) = $$(2)^3$$ – 4$$(2)^2$$ + 5(2) – 2 = 8 – 16 + 10 – 2 = 0

p(1) = $$(1)^3$$ – 4$$(1)^2 + 5(1) – 2 = 1 – 4 + 5 – 2 = 0 \(\therefore$$   2, 1 and 1 are the zeroes of $$x^3 – 4x^2 + 5x – 2$$

So,  $$\alpha$$ = 2, $$\beta$$ = 1 and $$\gamma$$ = 1.

Therefore, $$\alpha$$ + $$\beta$$ + $$\gamma$$ = 2 + 1 + 1 = $$-(-4)\over 1$$ = $$-b\over a$$

$$\alpha$$$$\beta$$ + $$\beta$$$$\gamma$$ + $$\gamma$$$$\alpha$$  = (2)(1) + (1)(1) + (1)(2) = $$5\over 1$$ = $$c\over a$$

and  $$\alpha$$$$\beta$$$$\gamma$$ = 2 $$\times$$ 1 $$\times$$ 1 = -(-2) = $$-d\over a$$