\( Ax^2+ Bx+ C \) Üç Terimlisi
1. Son Terimden Faydalanarak Çarpanlara Ayırma:
\( A = 1 \) ise çarpımları \( C \) ye, toplamları \( B \) ye eşit olan iki sayı araştırılır.
\[
B = m + n \quad \text{ve} \quad C = m \cdot n \quad \text{olmak üzere,}
\]
\[
x^2 + Bx + C = (x + m)(x + n) \quad \text{dir. }
\]
Örnekler:
\( \bullet \quad x^2 – 5x – 50 = (x – 10)(x + 5) \)
\[
\begin{array}{c c c}
= x^2 &- \; 5x \; & \; -50 \\
&\swarrow \searrow & \swarrow \searrow \\
& -10 +5 & -10 \cdot 5
\end{array}
\]
\( \bullet \quad x^2 – (a + 2)x + 2a = (x – 2)(x – a) \)
\[
\begin{array}{c c c}
= x^2 &- \; (a+2) x \; & \; +2 \cdot a \\
&\swarrow \searrow & \swarrow \searrow \\
& a + a^2 & a \cdot a^2
\end{array}
\]
\( \bullet \quad x^2 + (a + a^2)x + a^3 = (x + a)(x + a^2) \)
\[
\begin{array}{c c c}
= x^2 &- \; (a + a^2)x \; & \; + a^3 \\
&\swarrow \searrow & \swarrow \searrow \\
& a + a^2 & a. \cdot a^2
\end{array}
\]
\( \bullet \quad x^2 + 2ax – 3a^2 = (x – a)(x + 3a) \)
\[
\begin{array}{c c c}
= x^2 &- \; 2ax \; & \; -3 a^2 \\
&\swarrow \searrow & \swarrow \searrow \\
& -a + 3a & -a. \cdot 3a
\end{array}
\]
\( \bullet \quad 4^x – 2^{x+1} – 3 = (2^x)^2 – 2 \cdot 2^x – 3 \)
\[2^x = t \quad \text{dönüşümü yapılırsa} \]
\[
\begin{array}{c c c}
(2^x)^2- 2 \cdot 2^x – 3 = t^2 &- \; 2t \; & \; -3 \\
&\swarrow \searrow & \swarrow \searrow \\
& -3+1 & -3 \cdot 1
\end{array}
\]
\( \bullet \quad x^6 + 10x^3 + 21 = (x^3)^2 + 10x^3 + 21 \)
\[
x^3 = t \quad \text{dönüşümü yapılırsa,}
\]
\[
\begin{array}{c c c}
x^6 + 10x^3 + 21 = t^2 &+ \; 10t \; & \; +21 \\
&\swarrow \searrow & \swarrow \searrow \\
& 3+7 & 3 \cdot 7
\end{array}
\]
\( \bullet \quad x – 5\sqrt{x} + 6 = (\sqrt{x})^2 – 5\sqrt{x} + 6 \)
\[ \sqrt{x} = t \quad \text{dönüşümü yapılırsa,} \]
\[
\begin{array}{c c c}
(\sqrt{ x})^2 – 5\sqrt{x} + 6 = t^2 &- \; 5t \; & \; +6 \\
&\swarrow \searrow & \swarrow \searrow \\
& -2+ (-3) & -2 \cdot (-3)
\end{array}
\]
\[= (\sqrt{x } -2 ) ( \sqrt{x } -3 ) \]
2. İlk ve Son Terimden Faydalanarak Çarpanlara Ayırma:
\( A \neq 1 \) ise \( A = p.q \), \( C = m.n \) ve \( B = p.n + q.m \) olacak şekilde dört sayı araştırılır.
\[
Ax^2 + Bx + C = (px + m)(qx + n)
\]
\[
\begin{array}{c c c}
= Ax^2 &+ \; Bx \; & \; +C \\
\Downarrow & \Uparrow & \Downarrow \\
px& & m \\
&\searrow \nearrow \\
qx& & n \\
\hline
&px \cdot n + qx \cdot m = Bx
\end{array}
\]
dir.
Örnekler:
\( \bullet \quad 6x^2 + 19x + 15 = (3x + 5)(2x + 3) \)
\[
\begin{array}{c c c}
= 6x^2 &+ \; 19x \; & \; +15 \\
\Downarrow & \Uparrow & \Downarrow \\
3x& & 5 \\
&\searrow \nearrow \\
2x& & 3 \\
\hline
&3x \cdot 3 + 2x \cdot 5 = 19x \quad \text{ortadaki terim}
\end{array}
\]
\( \bullet \quad 6x^2 + 11x – 2 = (6x – 1)(x + 2) \)
\[
\begin{array}{c c c}
= 6x^2 &+ \; 11x \; & \; -2 \\
\Downarrow & \Uparrow & \Downarrow \\
6x& & -1 \\
&\searrow \nearrow \\
x& & 2 \\
\hline
&6x \cdot 2 + 2x \cdot (-1) = 11x \quad \text{ortadaki terim}
\end{array}
\]
\( \bullet \quad a^2 x^2 + (ab – a)x – b = (ax – 1)(ax + b) \)
\[
\begin{array}{c c c}
= a^2x^2 &+ \; (ab-a)x \; & \; -b \\
\Downarrow & \Uparrow & \Downarrow \\
ax& & -1 \\
&\searrow \nearrow \\
ax& & b \\
\hline
&ax \cdot b + ax \cdot (-1) = (ab-a)x \quad \text{ortadaki terim}
\end{array}
\]
\( \bullet \quad 6a^2 + ab – 2b^2 = (2a – b)(3a + 2b) \)
\[
\begin{array}{c c c}
= 6a^2 &+ \; ab \; & \; -2b^2 \\
\Downarrow & \Uparrow & \Downarrow \\
2a& & -b \\
&\searrow \nearrow \\
3a& & 2b \\
\hline
&2a \cdot 2b + 3a \cdot (-b) = ab \quad \text{ortadaki terim}
\end{array}
\]
\( \bullet \quad a^{2x+1} + (a+2)a^x + 2 = a(a^x)^2 + (a+2)a^x + 2 \)
\[ a^x = t \quad \text{dönüşümü yapılırsa } \]
\[ a(a^x)^2+ ( a+ 2) a^x +2 \]
\[
\begin{array}{c c c}
= at^2 &+ \; (a+2)t \; & \; 2 \\
\Downarrow & \Uparrow & \Downarrow \\
at& & 2 \\
&\searrow \nearrow \\
t& & 1 \\
\hline
&at \cdot 1 + t \cdot 2 = (a+2) t \quad \text{ortadaki terim}
\end{array}
\]
\[ a^{2x+1} + (a+2)a^x + 2 = (a^x+1 ) (a^{x+1}+ 2 ) \]
\( \bullet \quad 2x – \sqrt{x} – 1 = (2\sqrt{x} + 1)(\sqrt{x} – 1) \)
\[
\begin{array}{c c c}
2x &- \; \sqrt{x } \; & \; -1 \\
\Downarrow & \Uparrow & \Downarrow \\
2\sqrt{x } & & 1 \\
&\searrow \nearrow \\
\sqrt{x } & & -1 \\
\hline
& 2\sqrt{x } \cdot (-1) + \sqrt{x } \cdot 1 = – \sqrt{x } \quad \text{ortadaki terim}
\end{array}
\]
\( \bullet \quad 3x^6 + 4x^3 + 1 = (3x^3 + 1)(x^3 + 1) \)
\[
= (3x^3 + 1)(x+1)(x^2 – x + 1) \text{ dir.}
\]
\[
\begin{array}{c c c}
3x^3 &+ \; 4x^3 \; & \; 1 \\
\Downarrow & \Uparrow & \Downarrow \\
3x^3& & 1 \\
&\searrow \nearrow \\
x^3& & 1 \\
\hline
& 3 x^3\cdot (1) + x^3 \cdot 1 = 4x^3 \quad \text{ortadaki terim}
\end{array}
\]