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Verify that 5,-2 and \frac{1}{3} are the zeroes of the cubic polynomial p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right) and verify the relation between its zeroes and coefficients.
CBSE | Class 10 | Excercise 2B | Maths | Polynomials | RS Aggarwal
Verify that 5,-2 and \frac{1}{3} are the zeroes of the cubic polynomial p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right) and verify the relation between its zeroes and coefficients.
p(x)=3x3–10x2–27x+10
p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right)
p(5)=3×53–10×52–27×5+10=(375–250–135+10)=0
p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times 5+10\right)=(375-250-135+10)=0
p(–2)=3×–23–10×–22–27×(–2)+10=(–24–40+54+10)=0
p(-2)=\left[3 \times\left(-2^{3}\right)-10 \times\left(-2^{2}\right)-27 \times(-2)+10\right]=(-24-40+54+10)=0
p13=3×133–10×132–27×13+10=3×127–10×19–9+10
\mathrm{p}\left(\frac{1}{3}\right)=\left\{3 \times\left(\frac{1}{3}\right)^{3}-10 \times\left(\frac{1}{3}\right)^{2}-27 \times \frac{1}{3}+10\right\}=\left(3 \times \frac{1}{27}-10 \times \frac{1}{9}-9+10\right)
=19–109+1=1–10–99=09=0
=\left(\frac{1}{9}-\frac{10}{9}+1\right)=\left(\frac{1-10-9}{9}\right)=\left(\frac{0}{9}\right)=0
∴5,–2 and 13 are the zeroes of p(x). 
\therefore 5,-2 \text { and } \frac{1}{3} \text { are the zeroes of } p(x) \text {. }

Let \alpha=5, \beta=-2 and \gamma=\frac{1}{3}. Then we have:

(\alpha+\beta+\gamma)=\left(5-2+\frac{1}{3}\right)=\frac{10}{3}=\frac{-\left(\text { coef ficient of } x^{2}\right)}{\left(\text { coefficient of } x^{3}\right)}

(\alpha \beta+\beta \gamma+\gamma \alpha)=\left(-10-\frac{2}{3}+\frac{5}{3}\right)=\frac{-27}{3}=\frac{\text { coefficient of } x}{\text { coefficient of } x^{3}}

\alpha \beta \gamma=\left\{5 \times(-2) \times \frac{1}{3}\right\}=\frac{-10}{3}=\frac{-(\text { constant term })}{\left(\text { coef ficient of } x^{3}\right)}

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