Question

Write polynomial P as a product of linear factors: $ P(x)={{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x-15 $ .
[Hint 1 and -3 are zeros of P]

Answer 1

Hint: As given in the hint that 1 and -3 are zeros of given polynomial it means that $ \left( x-1 \right)\And \left( x+3 \right) $ are two factors of the polynomial so we will divide the given polynomial by $ \left( x-1 \right)\left( x+3 \right) $ to obtain other two factors.

Complete step by step answer:
We have been given a polynomial $ P(x)={{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x-15 $ .
We have to find the linear factors of the given polynomial.
We have given a hint in the question that 1 and -3 are zeros of a given polynomial so $ \left( x-1 \right)\And \left( x+3 \right) $ are two factors of the polynomial.
Now, to find the other factors we will divide the given polynomial $ P(x)={{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x-15 $ by $ \left( x-1 \right)\left( x+3 \right) $
Let us divide the polynomial by using long division method, we have
  $ \Rightarrow \left( x-1 \right)\left( x+3 \right)={{x}^{2}}+2x-3 $
  $ {{x}^{2}}+2x-3\overset{{{x}^{2}}+2x+5}{\overline{\left){\begin{align}
  & {{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x-15 \\
 & \dfrac{{{x}^{4}}+2{{x}^{3}}-3{{x}^{2}}}{\begin{align}
  & 2{{x}^{3}}+9{{x}^{2}}+4x \\
 & \dfrac{2{{x}^{3}}+4{{x}^{2}}-6x}{\begin{align}
  & 5{{x}^{2}}+10x-15 \\
 & 5{{x}^{2}}+10x-15 \\
 & 0 \\
\end{align}} \\
\end{align}} \\
\end{align}}\right.}} $
On dividing, we get quotient $ {{x}^{2}}+2x+5 $ and remainder zero.
Now we factorize the obtained quotient by using quadratic formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
We get
  $ \begin{align}
  & \Rightarrow x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times 5}}{2} \\
 & \Rightarrow x=\dfrac{-2\pm \sqrt{4-20}}{2} \\
 & \Rightarrow x=\dfrac{-2\pm \sqrt{-16}}{2} \\
 & \Rightarrow x=\dfrac{-2\pm 4i}{2} \\
 & \Rightarrow x=\dfrac{2\left( -1\pm 2i \right)}{2} \\
 & \therefore x=-1\pm 2i \\
\end{align} $
So, the two factors will be $ -1-2i $ , $ -1+2i $
So, the product of linear factors of $ P(x)={{x}^{4}}+4{{x}^{3}}+6{{x}^{2}}+4x-15 $ will be $ \left( x-1 \right)\left( x+3 \right)\left( x+1+2i \right)\left( x+1-2i \right) $ .

Note:
 The fact is that if $ a $ is the zero of the polynomial $ P(x) $ then $ (x-a) $ is the factor the polynomial. Also in this question, the highest power of $ x $ is 4 i.e. $ {{x}^{4}} $ that is the polynomial is of order 4 and it will have four zeros that mean four factors. Also in this question, we will use the quadratic formula to factorize the quadratic equation. Students also use the alternative method of factorization to find the factors of the quadratic equation.