The Quadratic Trinomial $Ax^2 + Bx + C$

 

1. Factoring by Analyzing the Constant Term ($A = 1$):

 

When $A = 1$, we look for two numbers whose product equals the constant term $C$, and whose sum equals the linear coefficient $B$.

\[
\text{Let } B = m + n \quad \text{and} \quad C = m \cdot n.
\]

\[
\text{Then, } x^2 + Bx + C = (x + m)(x + n).
\]

 

Examples:

 

\( \bullet \quad x^2 – 5x – 50 = (x – 10)(x + 5) \)

\[
\begin{array}{c c c}
= x^2 &- \; 5x \; & \; -50 \\
&\swarrow \searrow & \swarrow \searrow \\
& -10 + 5 & -10 \cdot 5
\end{array}
\]

\( \bullet \quad x^2 – (a + 2)x + 2a = (x – 2)(x – a) \)

\[
\begin{array}{c c c}
= x^2 &- \; (a+2) x \; & \; +2 \cdot a \\
&\swarrow \searrow & \swarrow \searrow \\
& -2 + (-a) & (-2) \cdot (-a)
\end{array}
\]

 

\( \bullet \quad x^2 + (a + a^2)x + a^3 = (x + a)(x + a^2) \)

\[
\begin{array}{c c c}
= x^2 &+ \; (a + a^2)x \; & \; + a^3 \\
&\swarrow \searrow & \swarrow \searrow \\
& a + a^2 & a \cdot a^2
\end{array}
\]

 

\( \bullet \quad x^2 + 2ax – 3a^2 = (x – a)(x + 3a) \)

 

\[
\begin{array}{c c c}
= x^2 &+ \; 2ax \; & \; -3 a^2 \\
&\swarrow \searrow & \swarrow \searrow \\
& -a + 3a & -a \cdot 3a
\end{array}
\]

 

\( \bullet \quad 4^x – 2^{x+1} – 3 = (2^x)^2 – 2 \cdot 2^x – 3 \)

\[ \text{Using substitution, let } 2^x = t: \]

\[
\begin{array}{c c c}
(2^x)^2 – 2 \cdot 2^x – 3 = t^2 &- \; 2t \; & \; -3 \\
&\swarrow \searrow & \swarrow \searrow \\
& -3 + 1 & -3 \cdot 1
\end{array}
\]
\[ = (t – 3)(t + 1) = (2^x – 3)(2^x + 1) \]

 

\( \bullet \quad x^6 + 10x^3 + 21 = (x^3)^2 + 10x^3 + 21 \)

\[ \text{Using substitution, let } x^3 = t: \]

\[
\begin{array}{c c c}
x^6 + 10x^3 + 21 = t^2 &+ \; 10t \; & \; +21 \\
&\swarrow \searrow & \swarrow \searrow \\
& 3 + 7 & 3 \cdot 7
\end{array}
\]
\[ = (t + 3)(t + 7) = (x^3 + 3)(x^3 + 7) \]

 

\( \bullet \quad x – 5\sqrt{x} + 6 = (\sqrt{x})^2 – 5\sqrt{x} + 6 \)

\[ \text{Using substitution, let } \sqrt{x} = t: \]

\[
\begin{array}{c c c}
(\sqrt{ x})^2 – 5\sqrt{x} + 6 = t^2 &- \; 5t \; & \; +6 \\
&\swarrow \searrow & \swarrow \searrow \\
& -2 + (-3) & -2 \cdot (-3)
\end{array}
\]

\[ = (t – 2)(t – 3) = (\sqrt{x} – 2)(\sqrt{x} – 3) \]

 

2. Factoring by Analyzing the Leading and Constant Terms ($A \neq 1$):

 

When $A \neq 1$, we look for four integers $p, q, m,$ and $n$ such that $A = p \cdot q$, $C = m \cdot n$, and the middle coefficient matches the sum of the cross-products: $B = p \cdot n + q \cdot m$.

\[
Ax^2 + Bx + C = (px + m)(qx + n)
\]

\[
\begin{array}{c c c}
= Ax^2 &+ \; Bx \; & \; +C \\
\Downarrow & \Uparrow & \Downarrow \\
px & & m \\
&\searrow \nearrow \\
qx & & n \\
\hline
& \text{Verify cross-products: } px \cdot n + qx \cdot m = Bx
\end{array}
\]

 

Examples:

 

\( \bullet \quad 6x^2 + 19x + 15 = (3x + 5)(2x + 3) \)

\[
\begin{array}{c c c}
= 6x^2 &+ \; 19x \; & \; +15 \\
\Downarrow & \Uparrow & \Downarrow \\
3x & & 5 \\
&\searrow \nearrow \\
2x & & 3 \\
\hline
& 3x \cdot 3 + 2x \cdot 5 = 19x \quad \text{(Matches middle term)}
\end{array}
\]

 

 

\( \bullet \quad 6x^2 + 11x – 2 = (6x – 1)(x + 2) \)

\[
\begin{array}{c c c}
= 6x^2 &+ \; 11x \; & \; -2 \\
\Downarrow & \Uparrow & \Downarrow \\
6x & & -1 \\
&\searrow \nearrow \\
x & & 2 \\
\hline
& 6x \cdot 2 + x \cdot (-1) = 11x \quad \text{(Matches middle term)}
\end{array}
\]

 

 

\( \bullet \quad a^2 x^2 + (ab – a)x – b = (ax – 1)(ax + b) \)

\[
\begin{array}{c c c}
= a^2x^2 &+ \; (ab-a)x \; & \; -b \\
\Downarrow & \Uparrow & \Downarrow \\
ax & & -1 \\
&\searrow \nearrow \\
ax & & b \\
\hline
& ax \cdot b + ax \cdot (-1) = (ab-a)x \quad \text{(Matches middle term)}
\end{array}
\]

 

 

\( \bullet \quad 6a^2 + ab – 2b^2 = (2a – b)(3a + 2b) \)

\[
\begin{array}{c c c}
= 6a^2 &+ \; ab \; & \; -2b^2 \\
\Downarrow & \Uparrow & \Downarrow \\
2a & & -b \\
&\searrow \nearrow \\
3a & & 2b \\
\hline
& 2a \cdot 2b + 3a \cdot (-b) = ab \quad \text{(Matches middle term)}
\end{array}
\]

 

 

\( \bullet \quad a^{2x+1} + (a+2)a^x + 2 = a(a^x)^2 + (a+2)a^x + 2 \)

\[ \text{Using substitution, let } a^x = t: \]

\[
\begin{array}{c c c}
= at^2 &+ \; (a+2)t \; & \; 2 \\
\Downarrow & \Uparrow & \Downarrow \\
at & & 2 \\
&\searrow \nearrow \\
t & & 1 \\
\hline
& at \cdot 1 + t \cdot 2 = (a+2)t \quad \text{(Matches middle term)}
\end{array}
\]
\[ \text{Factored form: } (at + 2)(t + 1) = (a \cdot a^x + 2)(a^x + 1) = (a^{x+1} + 2)(a^x + 1). \]

 

 

\( \bullet \quad 2x – \sqrt{x} – 1 = (2\sqrt{x} + 1)(\sqrt{x} – 1) \)

\[
\begin{array}{c c c}
2x &- \; \sqrt{x} \; & \; -1 \\
\Downarrow & \Uparrow & \Downarrow \\
2\sqrt{x} & & 1 \\
&\searrow \nearrow \\
\sqrt{x} & & -1 \\
\hline
& 2\sqrt{x} \cdot (-1) + \sqrt{x} \cdot 1 = -\sqrt{x} \quad \text{(Matches middle term)}
\end{array}
\]

 

 

\( \bullet \quad 3x^6 + 4x^3 + 1 = (3x^3 + 1)(x^3 + 1) \)

\[ \text{Factoring the sum of cubes } (x^3 + 1) \text{ completely yields:} \]
\[
= (3x^3 + 1)(x + 1)(x^2 – x + 1).
\]

\[
\begin{array}{c c c}
3x^6 &+ \; 4x^3 \; & \; 1 \\
\Downarrow & \Uparrow & \Downarrow \\
3x^3 & & 1 \\
&\searrow \nearrow \\
x^3 & & 1 \\
\hline
& 3x^3 \cdot 1 + x^3 \cdot 1 = 4x^3 \quad \text{(Matches middle term)}
\end{array}
\]

 

 

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