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Question: Find all zeros of the polynomial \(2{{x}^{4}}+7{{x}^{3}}-19{{x}^{2}}-14x+30\), if two of its zeros a...

Find all zeros of the polynomial 2x4+7x3−19x2−14x+302{{x}^{4}}+7{{x}^{3}}-19{{x}^{2}}-14x+30, if two of its zeros are 2\sqrt{2} and −2-\sqrt{2}

Explanation

Solution

Hint:We will use the fact that if “a” is a zero of the polynomial p(x), then (x – a) will be a factor of the polynomial p(x). So, we will divide the given polynomial by (x−2)(x+2)\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right) . The quotient will be a quadratic in x. We will solve the quadratic to find the other zeroes.

Complete step-by-step answer:
In a given question, we have a polynomial 2x4+7x3+−19x2−14x+302{{x}^{4}}+7{{x}^{3}}+-19{{x}^{2}}-14x+30. Here, the highest power of xx is 44, that is the polynomial is of order 44. So, it will have four zeros. Also 2 of the zeros of this polynomial is given in question, which are 2\sqrt{2} and −2-\sqrt{2}. Let us consider, the remaining two zeros of this polynomial to be α\alpha and β\beta
Therefore, this polynomial will be product of factors (x−2), (x−(−2)), (x−α), (x−β)\left( x-\sqrt{2} \right),\,\left( x-\left( -\sqrt{2} \right) \right),\,\left( x-\alpha \right),\,\left( x-\beta \right).
Hence, we can write, (taking kk as any constant).
k(x−2)(x−(−2))(x−α)(x−β)=2x4+7x3−19x2−14x+30 ⇒k(x−2)(x+2)(x−α)(x−β)=2x4+7x3−19x2−14x+30 \begin{aligned} & k\left( x-\sqrt{2} \right)\left( x-\left( -\sqrt{2} \right) \right)\left( x-\alpha \right)\left( x-\beta \right)=2{{x}^{4}}+7{{x}^{3}}-19{{x}^{2}}-14x+30 \\\ & \Rightarrow k\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)\left( x-\alpha \right)\left( x-\beta \right)=2{{x}^{4}}+7{{x}^{3}}-19{{x}^{2}}-14x+30 \\\ \end{aligned}
Applying formula (a−b)(a+b)=(a2+b2)\left( a-b \right)\left( a+b \right)=\left( {{a}^{2}}+{{b}^{2}} \right) we get,
k(x2−(2)2)(x−α)(x−β)=2x4+7x3+−19x2−14x+30 ⇒k(x2−2)(x−α)(x−β)=2x4+7x3+−19x2−14x+30 \begin{aligned} & k\left( {{x}^{2}}-{{\left( \sqrt{2} \right)}^{2}} \right)\left( x-\alpha \right)\left( x-\beta \right)=2{{x}^{4}}+7{{x}^{3}}+-19{{x}^{2}}-14x+30 \\\ & \Rightarrow k\left( {{x}^{2}}-2 \right)\left( x-\alpha \right)\left( x-\beta \right)=2{{x}^{4}}+7{{x}^{3}}+-19{{x}^{2}}-14x+30 \\\ \end{aligned}
Dividing (x2−2)\left( {{x}^{2}}-2 \right) from both sides of the equation, we get,
k(x−α)(x−β)=2x4+7x3−19x2−14x+30x2−2k\left( x-\alpha \right)\left( x-\beta \right)=\dfrac{2{{x}^{4}}+7{{x}^{3}}-19{{x}^{2}}-14x+30}{{{x}^{2}}-2}
Let us divide right hand side of this equation using a long division method.

& {{x}^{2}}-2\overset{2{{x}^{2}}+7x-15}{\overline{\left){2{{x}^{4}}+7{{x}^{3}}-19{{x}^{2}}-14x+30}\right.}} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{x}^{4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-14x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7{{x}^{3}}-15{{x}^{2}}-14x}+30 \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7{{x}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-14x \\\ & \,\,\,\,\,\,\,\,\,\,\overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-15{{x}^{2}}+0+30}\,\, \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-15{{x}^{2}}\,\,\,\,\,\,\,\,\,+30 \\\ & \,\,\,\,\,\,\,\,\,\,\,\overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\ \end{aligned}$$ So, on dividing, we get quotient $2{{x}^{2}}+7x-15$ and remainder $0$. Using this value in equation(i), we get, $k\left( x-\alpha \right)\left( x-\beta \right)=2{{x}^{2}}+7x-15$ Splitting the middle term of left-hand side of above equation to factorise it, we get, $k\left( x-\alpha \right)\left( x-\beta \right)=2{{x}^{2}}+10x-3x-15$ Taking common terms out, we get, $k\left( x-\alpha \right)\left( x-\beta \right)=2x\left( x+5 \right)-3x\left( x+5 \right)$ $\,\,=\left( 2x-3 \right)\left( x+5 \right)$ Taking 2 common from first factor, we get, $k\left( x-\alpha \right)\left( x-\beta \right)=2\left( x-\dfrac{3}{2} \right)\left( x-\left( 5 \right) \right)$ Comparing both side of equation, we get, $k=2,\,\alpha =\dfrac{3}{2},\,\beta =-5$ Hence, all zeros of polynomial are, $\sqrt{2},\,-\sqrt{2},\,\dfrac{3}{2},\,-5$ Note: While doing calculations, be careful of the sign. Sign mistakes are very common and hence, students should be very careful while doing calculations and substitutions.Students should also know about the calculation procedure of long division method for solving these types of questions.