Question

Find the zeroes of the polynomial f(x) =2x³– 11x² +
10x + 8, if the product of its two zeroes is 8.​

Answer 1

Concept to be used:

• $\sf{\alpha \beta \gamma=-\dfrac{d}{a} }$

→ [solutions of $\sf{ax^3+bx^2+cx+d=0}$ ]

• Factor Theorem

Solving the Problem

(from the above)

• The product of all zeros is -4.

(given that)

• The product of the two zeros is 8.

∴One zero should be -1/2.

∴x=-1/2

From the zero, we know that x+1/2 a factor.

If we divide by x+1/2, we get other factors.

$\sf{f(x)=(x+\dfrac{1}{2} )(2x^2-12x+16)}$

$\sf{f(x)=2(x+\dfrac{1}{2} )(x-2)(x-4)}$

Conclusion

The three zeros are -1/2, 2, 4.

For your information,

1. The first concept is the relation between roots and coefficients.

2. We divide by x+1/2 according to the factor theorem.