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The polynomial $2x^3 + 6x^2 - 7x - 10$ has zeros $\alpha$, $\beta$ and $\gamma$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2018 - Paper 1

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Question 4

The-polynomial-$2x^3-+-6x^2---7x---10$-has-zeros-$\alpha$,-$\beta$-and-$\gamma$-HSC-SSCE Mathematics Extension 1-Question 4-2018-Paper 1.png

The polynomial $2x^3 + 6x^2 - 7x - 10$ has zeros $\alpha$, $\beta$ and $\gamma$. What is the value of $\alpha\beta\gamma(\alpha + \beta + \gamma)$?

Worked Solution & Example Answer:The polynomial $2x^3 + 6x^2 - 7x - 10$ has zeros $\alpha$, $\beta$ and $\gamma$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2018 - Paper 1

Step 1

Find the sum of the roots, $\alpha + \beta + \gamma$

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Answer

Using Vieta's formulas, the sum of the roots for the polynomial 2x3+6x2−7x−102x^3 + 6x^2 - 7x - 10 is given by −62=−3-\frac{6}{2} = -3. Therefore, α+β+γ=−3\alpha + \beta + \gamma = -3.

Step 2

Find the product of the roots, $\alpha \beta \gamma$

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Answer

Similarly, the product of the roots is given by −−102=5-\frac{-10}{2} = 5. Thus, αβγ=5\alpha \beta \gamma = 5.

Step 3

Calculate $\alpha \beta \gamma (\alpha + \beta + \gamma)$

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Answer

Now, substituting the values obtained:

αβγ(α+β+γ)=5⋅(−3)=−15.\alpha \beta \gamma (\alpha + \beta + \gamma) = 5 \cdot (-3) = -15.

Thus, the final value is −15-15.

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