Question

Find all the zeroes of 2x^4-3x^3-3x^2+6x-2 if you know that two of the zeroes are 1 and 1/2

Answer 1

                   

Answer: √2 and -√2

     

Here, 1 and 1/2 are the known zeroes.

So (x - 1) and (x - 1/2)  are two factors of p(x) = 2x⁴-3x³-3x²+6x-2 = 0.

First we have to multiply the known factors.

$(x-1)(x-\frac{1}{2}) \\ \\ \Rightarrow\ x^2-\frac{1}{2}x-x+\frac{1}{2} \\ \\ \Rightarrow\ x^2-\frac{3}{2}x+\frac{1}{2}$

Let this be q(x).

q(x) = x² - 3/2 x + 1/2

Now we've to divide p(x) by q(x).

$2x^4-3x^3-3x^2+6x-2 \\ \\ \Rightarrow\ 2x^4-3x^3+x^2-4x^2+6x-2 \\ \\ \Rightarrow\ 2x^2(x^2-\frac{3}{2}x+\frac{1}{2})-4(x^2-\frac{3}{2}x+\frac{1}{2}) \\ \\ \Rightarrow\ (x^2-\frac{3}{2}x+\frac{1}{2})(2x^2-4)$

Here, we get the quotient 2x² - 4. This is the product of the other zeroes.

Let's find the other zeroes.

$2x^2-4=0 \\ \\ \Rightarrow\ 2x^2=4 \\ \\ \Rightarrow\ x^2=2 \\ \\ \Rightarrow\ x=\bold{\pm \sqrt{2}}$

∴ √2 and -√2 are the other zeroes.

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Answer 2

Answer:

Here, 1 and 1/2 are the known zeroes.

So (x - 1) and (x - 1/2)  are two factors of p(x) = 2x⁴-3x³-3x²+6x-2 = 0.

First we have to multiply the known factors.

\begin{gathered}(x-1)(x-\frac{1}{2}) \\ \\ \Rightarrow\ x^2-\frac{1}{2}x-x+\frac{1}{2} \\ \\ \Rightarrow\ x^2-\frac{3}{2}x+\frac{1}{2}\end{gathered}(x−1)(x−21)⇒ x2−21x−x+21⇒ x2−23x+21

Let this be q(x).

q(x) = x² - 3/2 x + 1/2

Now we've to divide p(x) by q(x).

\begin{gathered}2x^4-3x^3-3x^2+6x-2 \\ \\ \Rightarrow\ 2x^4-3x^3+x^2-4x^2+6x-2 \\ \\ \Rightarrow\ 2x^2(x^2-\frac{3}{2}x+\frac{1}{2})-4(x^2-\frac{3}{2}x+\frac{1}{2}) \\ \\ \Rightarrow\ (x^2-\frac{3}{2}x+\frac{1}{2})(2x^2-4) \\ \\\end{gathered}2x4−3x3−3x2+6x−2⇒ 2x4−3x3+x2−4x2+6x−2⇒ 2x2(x2−23x+21)−4(x2−23x+21)⇒ (x2−23x+21)(2x2−4)

Here, we get the quotient 2x² - 4. This is the product of the other zeroes.

Let's find the other zeroes.

\begin{gathered}2x^2-4=0 \\ \\ \Rightarrow\ 2x^2=4 \\ \\ \Rightarrow\ x^2=2 \\ \\ \Rightarrow\ x=\bold{\pm \sqrt{2}}\end{gathered}2x2−4=0⇒ 2x2=4⇒ x2=2⇒ x=±2

∴ √2 and -√2 are the other zeroes.

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