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It is known that a quartic function with a leading coefficient of 1 can be factored into two quadratic functions according to this rule

It is known that a quartic function with a leading coefficient of 1 can be factored-example-1
User Hanfei Sun
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1 Answer

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ANSWER:

The five terms quadratic polynomial:


\begin{gathered} x^4+2x^3+3x^2+2x+1\operatorname{\rightarrow}\text{ expanded form} \\ \\ (x^2+x+1)(x^2+x+1)\operatorname{\rightarrow}\text{ factored form} \end{gathered}

The two terms quadratic polynomial:


\begin{gathered} \begin{equation*} x^2+5x+6\rightarrow\text{ expanded form} \end{equation*} \\ \\ (x+3)(x+2)\operatorname{\rightarrow}\text{ factored form} \end{gathered}

Explanation:

If we pick that:

p = 1

q = 1

r = 1

s = 1

So:


\begin{gathered} a=p+r=1+1=2 \\ \\ b=q+s+pr=1+1+1\cdot1=3 \\ \\ c=ps+qr=1\cdot1+1\cdot1=2 \\ \\ d=qs=1\cdot1=1 \end{gathered}

The five terms quadratic polynomial is:


\begin{gathered} x^4+ax^3+bx^2+cx+d=(x^2+px+q)(x^2+rx+s) \\ \\ \text{ we replacing} \\ \\ x^4+2x^3+3x^2+2x+1\rightarrow\text{ expanded form} \\ \\ \left(x^2+x+1\right)\left(x^2+x+1\right)\rightarrow\text{ factored form} \end{gathered}

The two terms quadratic polynomial is:


\begin{gathered} ax^2+(b+c)x+bc=(x+b)(x+c) \\ \\ \text{ we replacing} \\ \\ x^2+(3+2)x+3\cdot2\rightarrow x^2+5x+6\rightarrow\text{ expanded form} \\ \\ (x+3)(x+2)\rightarrow\text{ factored form} \end{gathered}

User DsD
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