Question

Verify that -1, 1/2, 1/3 are the zeroes of the polynomial 6x3 + x2 – 4x + 1 and verify the relationship between the zeroes and the coefficients.(A)

Answer 1

✳ Objective ✳

➩ Verify that -1, 1/2, 1/3 are the zeroes of the polynomial 6x³ + x² - 4x + 1.

➩ Verify the relationship between the zeroes and the coefficients.

✳ Solution ✳

Let p(x) = 6x³ + x² - 4x + 1.

When we put  -1, 1/2, 1/3 individually in the place of x, and if the result is 0, then these are the zeros of p(x).

_________________

$\displaystyle{\sf{p(-1)=6(-1)^3+(-1)^2-4(-1)+1 }}\\\\\displaystyle{\sf{ =-6+1+4+1}}\\\\\displaystyle{\sf{ =0}}$

Hence, -1 is a zero of p(x).

_________________

$\displaystyle{\sf{p\bigg(\frac{1}{2}\bigg)=6 \bigg(\frac{1}{2} \bigg)^3+\bigg(\frac{1}{2} \bigg)^2 -4 \bigg( \frac{1}{2} \bigg)+1}}\\\\\displaystyle{\sf{ =\frac{6}{8}+\frac{1}{4}-\frac{4}{2}+1 }}\\\\\displaystyle{\sf{ =\frac{6+2-16+8}{8} }}\\\\\displaystyle{\sf{ =\frac{0}{8} }}\\\\\displaystyle{\sf{ =0}}$

Hence, 1/2 is a zero of p(x).

_________________

$\displaystyle{\sf{p\bigg(\frac{1}{3}\bigg)=6 \bigg(\frac{1}{3} \bigg)^3+\bigg(\frac{1}{3} \bigg)^2 -4 \bigg( \frac{1}{3} \bigg)+1}}\\\\\displaystyle{\sf{ =\frac{6}{27}+\frac{1}{9}-\frac{4}{3}+1 }}\\\\\displaystyle{\sf{ =\frac{6+3-36+27}{27} }}\\\\\displaystyle{\sf{ =\frac{0}{27} }}\\\\\displaystyle{\sf{ =0}}$

Hence, 1/3 is a zero of p(x).

$\underline{\rule{180}2}$

Let α = -1, β = 1/2 and γ = 1/3.

In the given polynomial p(x), a = 6, b = 1, c = -4 and d = 1.

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$\sf{\alpha + \beta + \gamma = -1 + \dfrac{1}{2} + \dfrac{1}{3}}\\\\\sf{\implies \alpha + \beta + \gamma = \dfrac{-6+3+2}{6} }\\\\\sf{\implies \alpha + \beta + \gamma = \dfrac{-1}{6} }$

$\rule{80}1$

$\sf{\dfrac{-b}{a}=\dfrac{-(1)}{6}}\\\\\sf{\implies \dfrac{-b}{a}=\dfrac{-1}{6}}$

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$\sf{\alpha \beta \gamma=-1 \times \dfrac{1}{2}\times \dfrac{1}{3}}\\\\\sf{\implies \alpha \beta \gamma=\dfrac{-1}{6}}$

$\rule{80}1$

$\sf{\dfrac{-d}{a}=\dfrac{-(1)}{6}}\\\\\sf{\implies \dfrac{-d}{a}=\dfrac{-1}{6}}$

__________________

$\sf{ \alpha \beta+\beta \gamma + \gamma \alpha =[-1\times\frac{1}{2} ]+[\frac{1}{2}\times\frac{1}{3} ]+[\frac{1}{3} \times -1] }\\\\\sf{ \alpha \beta+\beta \gamma + \gamma \alpha = \frac{-1}{2}+\frac{1}{6}+\frac{-1}{3} }\\\\\sf{ \alpha \beta+\beta \gamma + \gamma \alpha = \frac{-3+1-2}{6} }\\\\\sf{ \alpha \beta+\beta \gamma + \gamma \alpha = \frac{-4}{6} }\\\\\sf{ \alpha \beta+\beta \gamma + \gamma \alpha = \frac{-2}{3} }$

$\rule{80}1$

$\sf{\dfrac{c}{a}=\dfrac{-4}{6}}\\\\\sf{\implies \dfrac{c}{a}=\dfrac{-2}{3}}$

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Relationship verified ✅

Answer 2

Correct Question:

Verify that -1, 1/2, 1/3 are the zeroes of the polynomial 6x³ + x² – 4x + 1 and verify the relationship between the zeroes and the coefficients.

SOLUTION:

Given :

Polynomial P(x) = 6x³ + x² – 4x + 1

Three number : (-1), (1/2) and (1/3)

To prove,

• Verify that -1, 1/2, 1/3 are the zeroes of the P(x) = 6x³ + x² - 4x + 1.
• Verify the relationship between the zeroes and the coefficients.

So,

P(-1) = 0 or not

P(-1) = 6(-1)³ + (-1)² - 4(-1) + 1.

       = 6 × (-1) + 1 + 4 + 1.

       = -6 + 1 + 4 + 1

       = -6 + 6

       = 0

P(-1) = 0, so, (-1) is the zero of polyniomial P(x).

$\rule{200}2$

$\bf P(\frac{1}{2}) = 0\ or\ not$

$\bf P(\frac{1}{2}) =6 (\frac{1}{2})^3+(\frac{1}{2} )^2 -4 (\frac{1}{2} )+1$

$\bf P( \frac{1}{2} ) = \frac{6}{8} + \frac{1}{4} - \frac{4}{2} + 1$

$\bf P(\frac{1}{2}) =\frac{6+2-16+8}{8}$

$\bf P(\frac{1}{2})=\frac{0}{8}$

$\bf P(\frac{1}{2})=0$

P(1/2) is also a zero of P(x).

$\rule{200}2$

$\bf P( \frac{1}{3} ) = 6(\frac{1}{3} )^3 + ( \frac{1}{3} )^2 - 4 ( \frac{1}{3} ) + 1$

$\bf P(\frac{1}{3}) =\frac{6}{27}+\frac{1}{9}-\frac{4}{3}+1$

$\bf P(\frac{1}{3}) = \frac{2}{9}+\frac{1}{9}-\frac{4}{3} +1$

$\bf P(\frac{1}{3}) =\frac{2+1}{9}-\frac{4}{3}+1$

$\bf P(\frac{1}{3}) = \frac{3}{9}- \frac{4}{3}+1$

$\bf P(\frac{1}{3}) = \frac{1}{3}-\frac{4}{3}+1$

$\bf P(\frac{1}{3})=\frac{1-4}{3} +1$

$\bf P(\frac{1}{3})=\frac{-3}{3}+1$

$\bf P(\frac{1}{3})=-1+1$

$\bf P(\frac{1}{3})=0$

So,

(1/3) is also a zero.

$\rule{200}1$

Now,

To verify the relationship between the zeroes and the coefficients.

From the polynomial, 6x³ + x² – 4x + 1

a = 6, b = 1, c = ( - 4 ), d = 1

And, we verified that (-1), (1/2), and (1/3) are zeroes so,

Let,

$\boxed{\red{\alpha = -1} ,\ \pink{\beta = \frac{1}{2}} ,\ \green{\gamma = \frac{1}{3}}}$

$\bf If\ \alpha,\beta,\gamma\ \bf are\ the\ zeroes\ of\ a\ cubic\ polynomial,\\ \bf ax^3+bx^2+cx+d,\ then$

$\boxed{\sf \alpha+\beta+\gamma =\frac{-b}{a}}$, $\boxed{\sf \alpha \beta+\beta \gamma + \gamma \alpha=\frac{c}{a}}$, $\boxed{\sf \alpha \beta \gamma =\frac{-d}{a}}$

So,

$\sf \dfrac{-b}{a}= \dfrac{-1}{6}\\ \\ \\ \sf \dfrac{c}{a} = \dfrac{-4}{6} = \dfrac{-2}{3} \\ \\ \\ \sf \dfrac{-d}{a} = \dfrac{-1}{6}$

$\star \rule{50}2 \star \rule{50}2 \star$

$\boxed{\sf \alpha+\beta+\gamma =\frac{-b}{a}} \\ \\ \\ \sf =(-1)+(\frac{1}{2})+\frac{1}{3} \\ \\ \\ \sf = \dfrac{-6+3+2}{6} \\ \\ \\ \sf = \dfrac{-6+5}{6}\\ \\ \\ = \boxed{\frac{-1}{6}} = \dfrac{-b}{a}$

$\rule{150}2$

$\boxed{\sf \alpha \beta+\beta \gamma + \gamma \alpha=\frac{c}{a}} \\ \\ \\ \sf= \bigg[(-1)\times (\frac{1}{2})\bigg] + \bigg[(\frac{1}{2}) \times (\frac{1}{3}) \bigg]+ \bigg[(\frac{1}{3})\times(-1)\bigg] \\ \\ \\ \sf = \frac{-1}{2} + \frac{1}{6} + \frac{-1}{3} \\ \\ \\ \sf = \dfrac{-3+1-2}{6} \\ \\ \\ \sf = \dfrac{-4}{6} \\ \\ \\ \sf =\boxed{ \frac{-2}{3}} = \dfrac{c}{a}$

$\rule{150}2$

$\boxed{\sf \alpha \beta \gamma =\frac{-d}{a}} \\ \\ \\ \sf = (-1) \times (\frac{1}{2}) \times \frac{1}{3} \\ \\ \\ \sf = \boxed{\frac{-1}{6}}=\dfrac{-d}{a}$

Hence,

The relationship between the zeroes and the coefficients. is verified.