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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 ax3+bx2+cx+dax^3 + bx^2 + cx + d is given by −ba-\frac{b}{a}. In this case, a=2a = 2 and b=6b = 6. Hence, α+β+γ=−62=−3\alpha + \beta + \gamma = -\frac{6}{2} = -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 −da-\frac{d}{a}. Here, d=−10d = -10, so αβγ=−−102=5\alpha \beta \gamma = -\frac{-10}{2} = 5.

Step 3

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

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Answer

Now we multiply these results: αβγ(α+β+γ)=5×−3=−15\alpha \beta \gamma (\alpha + \beta + \gamma) = 5 \times -3 = -15.

Step 4

Final Answer

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Answer

The value is −15-15, which corresponds to option B.

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