แบบฝึกหัด การแยกตัวประกอบของพหุนามดีกรีสอง

1. จงแยกตัวประกอบของพหุนามต่อไปนี้

\(2x^2 + 5x + 2\)
\(2x^2 + x \; – 6\)
\(3x^2 \;- 13x + 12\)
\(6x^2 + 5x \; – 25\)
\(3x^2 \; – x \; – 10\)
\(8x^2 + 2x \; – 15\)
\(4x^2 + 19x \; – 30\)
\(7x^2 + 39x \; – 18\)
\(5x^2 + 23x + 12\)
\(9x^2 + 9x \; – 40\)
\(6x^2 + 19x + 15\)
\(10x^2 + 29x \; – 21\)
\(2x^2 \; – 5x \; – 3\)
\(-5x^2 + 7x + 6\)
\(2x^2 \; – 7x \; – 15\)
\(-6x^2 +13x + 8\)
\(3x^2 \; – 17x \; – 6\)
\(-18x^2 \; – 9x + 2\)
\(6x^2 \; – x \; – 40\)
\(16x^2 + 66x \; – 27\)
\(7x^2 \; – 5x \; – 2\)
\(28x^2 + 31x \; – 5\)
\(8x^2 \; – 6x \; – 9\)
\(12x^2 \; – 29x + 14\)

ดูเฉลยคำตอบ

\(2x^2 + 5x + 2 = (2x + 1)(x + 2)\)
\(2x^2 + x \; – 6 = (2x \; – 3)(x + 2)\)
\(3x^2 \;- 13x + 12 = (3x \; – 4)(x \; – 3)\)
\(6x^2 + 5x \; – 25 = (3x \; – 5)(2x + 5)\)
\(3x^2 \; – x \; – 10 = (3x + 5)(x \; – 2)\)
\(8x^2 + 2x \; – 15 = (4x \; – 5)(2x + 3)\)
\(4x^2 + 19x \; – 30 = (4x \; – 5)(x + 6)\)
\(7x^2 + 39x \; – 18 = (7x \; – 3)(x + 6)\)
\(5x^2 + 23x + 12 = (5x + 3)(x + 4)\)
\(9x^2 + 9x \; – 40 = (3x + 8)(3x \; – 5)\)
\(6x^2 + 19x + 15 = (3x + 5)(2x + 3)\)
\(10x^2 + 29x \; – 21 = (5x \; – 3)(2x + 7)\)
\(2x^2 \; – 5x \; – 3 = (2x + 1)(x \; – 3)\)
\(-5x^2 + 7x + 6 = (-5x \; – 3)(x \; – 2)\)
\(2x^2 \; – 7x \; – 15 = (2x + 3)(x \; – 5)\)
\(-6x^2 +13x + 8 = (-3x + 8)(2x + 1)\)
\(3x^2 \; – 17x \; – 6 = (3x + 1)(x \; – 6)\)
\(-18x^2 \; – 9x + 2 = (6x \; – 1)(-3x \; – 2)\)
\(6x^2 \; – x \; – 40 = (3x \; – 8)(2x + 5)\)
\(16x^2 + 66x \; – 27 = (8x \; – 3)(2x + 9)\)
\(7x^2 \; – 5x \; – 2 = (7x + 2)(x \; – 1)\)
\(28x^2 + 31x \; – 5 = (7x \; – 1)(4x + 5)\)
\(8x^2 \; – 6x \; – 9 = (4x + 3)(2x \; – 3)\)
\(12x^2 \; – 29x + 14 = (3x \; – 2)(4x \; – 7)\)

2. จงแยกตัวประกอบของพหุนามต่อไปนี้ โดยวิธีทำเป็นกำลังสองสมบูรณ์

\(x^2 + 8x \; – 84\)
\(x^2 \; – 8x + 15\)
\(x^2 + 8x + 15\)
\(x^2 \; – 24x + 143\)
\(x^2 \; – 2x \; – 8\)
\(x^2 + 10x \; – 7\)
\(x^2 \; – 7x \; – 1\)
\(x^2 + 5x \; – 3\)
\(2x^2 + 3x + 1\)
\(2x^2 + x \; – 3\)

ดูเฉลยคำตอบ

\(x^2 + 8x \; – 84\)
\(\begin{align*}
\qquad &= x^2 + 2(4)x + (4)^2 \; – (4)^2 \;- 84 \\
&= (x + 4)^2 \; – 16 \; – 84 \\
&= (x + 4)^2 \; – 10^2 \\
&= (x + 4 + 10)(x + 4 \; – 10) \\
&= (x + 14)(x \; – 6)
\end{align*}\)
\(x^2 \; – 8x + 15\)
\(\begin{align*}
\qquad &= x^2 \; – 2(4)x + (4)^2 \; – (4)^2 + 15 \\
&= (x \; – 4)^2 \; – 16 + 15 \\
&= (x \; – 4)^2 \; – (1)^2 \\
&= (x \; – 4 + 1)(x \; – 4 \; – 1) \\
&= (x \; – 3)(x \; – 5)
\end{align*}\)
\(x^2 + 8x + 15\)
\(\begin{align*}
\qquad &= x^2 \; + 2(4)x + (4)^2 \; – (4)^2 + 15 \\
&= (x + 4)^2 \; – 16 + 15 \\
&= (x + 4)^2 \; – (1)^2 \\
&= (x + 4 + 1)(x + 4 \; – 1) \\
&= (x + 5)(x + 3)
\end{align*}\)
\(x^2 \; – 24x + 143\)
\(\begin{align*}
\qquad &= x^2 \; – 2(12)x + (12)^2 \; – (12)^2 + 143 \\
&= (x \; – 12)^2 \; – 144 + 143 \\
&= (x \; – 12)^2 \; – (1)^2 \\
&= (x \; – 12 + 1)(x \; – 12 \; – 1) \\
&= (x \; – 11)(x \; – 13)
\end{align*}\)
\(x^2 \; – 2x \; – 8\)
\(\begin{align*}
\qquad &= x^2 \; – 2(1)x + (1)^2 \; – (1)^2 \; – 8 \\
&= (x \; – 1)^2 \; – 1 \; – 8 \\
&= (x \; – 1)^2 \; – (3)^2 \\
&= (x \; – 1 + 3)(x \; – 1 \; – 3) \\
&= (x + 2)(x \; – 4)
\end{align*}\)
\(x^2 + 10x \; – 7\)
\(\begin{align*}
\qquad &= x^2 + 2(5)x + (5)^2 \; – (5)^2 \; – 7 \\
&= (x + 5)^2 \; – 25 \; – 7 \\
&= (x + 5)^2 \; – (\sqrt{32})^2 \\
&= (x + 5)^2 \; – (4\sqrt{2})^2 \\
&= (x + 5 + 4\sqrt{2})(x + 5 \; – 4\sqrt{2})
\end{align*}\)
\(x^2 \; – 7x \; – 1\)
\(\begin{align*}
\qquad &= x^2 \; – 2\Bigl(\frac{7}{2}\Bigl)x + \Bigl(\frac{7}{2}\Bigl)^2 \; – \Bigl(\frac{7}{2}\Bigl)^2 \; – 1 \\
&= \Bigl(x \; – \frac{7}{2}\Bigl)^2 \; – \frac{49}{4} \; – 1 \\
&= \Bigl(x \; – \frac{7}{2}\Bigl)^2 \; – \frac{53}{4} \\
&= \Bigl(x \; – \frac{7}{2}\Bigl)^2 \; – \Bigl(\frac{\sqrt{53}}{2}\Bigl)^2 \\
&= \Bigl(x \; – \frac{7 + \sqrt{53}}{2}\Bigl)\Bigl(x \; – \frac{7 \; – \sqrt{53}}{2}\Bigl)
\end{align*}\)
\(x^2 + 5x \; – 3\)
\(\begin{align*}
\qquad &= x^2 + 2\Bigl(\frac{5}{2}\Bigl)x + \Bigl(\frac{5}{2}\Bigl)^2 \; – \Bigl(\frac{5}{2}\Bigl)^2 \; – 3 \\
&= \Bigl(x + \frac{5}{2}\Bigl)^2 \; – \frac{25}{4} \; – 3 \\
&= \Bigl(x + \frac{5}{2}\Bigl)^2 \; – \frac{37}{4} \\
&= \Bigl(x + \frac{5}{2}\Bigl)^2 \; – \Bigl(\frac{\sqrt{37}}{2}\Bigl)^2 \\
&= \Bigl(x + \frac{5 + \sqrt{37}}{2}\Bigl)\Bigl(x + \frac{5 \; – \sqrt{37}}{2}\Bigl)
\end{align*}\)
\(2x^2 + 3x + 1\)
\(\begin{align*}
\qquad &= 2\biggl[x^2 + \frac{3}{2}x + \frac{1}{2}\biggl] \\
&= 2\biggl[x^2 + 2\Bigl(\frac{3}{4}\Bigl)x + \Bigl(\frac{3}{4}\Bigl)^2 \; – \Bigl(\frac{3}{4}\Bigl)^2 + \frac{1}{2}\biggl] \\
&= 2\biggl[\Bigl(x + \frac{3}{4}\Bigl)^2 \; – \frac{9}{16} + \frac{1}{2}\biggl] \\
&= 2\biggl[\Bigl(x + \frac{3}{4}\Bigl)^2 \; – \frac{1}{16}\biggl] \\
&= 2\biggl[\Bigl(x + \frac{3}{4}\Bigl)^2 \; – \Bigl(\frac{1}{4}\Bigl)^2\biggl] \\
&= 2\biggl[\Bigl(x + \frac{3}{4} + \frac{1}{4}\Bigl)\Bigl(x + \frac{3}{4} \; – \frac{1}{4}\Bigl)\biggl] \\
&= 2\biggl[(x + 1)\Bigl(x + \frac{1}{2}\Bigl)\biggl] \end{align*}\)
\(2x^2 + x \; – 3\)
\(\begin{align*}
\qquad &= 2\biggl[x^2 + \frac{1}{2}x \; – \frac{3}{2}\biggl] \\
&= 2\biggl[x^2 + 2\Bigl(\frac{1}{4}\Bigl)x + \Bigl(\frac{1}{4}\Bigl)^2 \; – \Bigl(\frac{1}{4}\Bigl)^2 \; – \frac{3}{2}\biggl] \\
&= 2\biggl[\Bigl(x + \frac{1}{4}\Bigl)^2 \; – \frac{1}{16} \; – \frac{3}{2}\biggl] \\
&= 2\biggl[\Bigl(x + \frac{1}{4}\Bigl)^2 \; – \frac{25}{16}\biggl] \\
&= 2\biggl[\Bigl(x + \frac{1}{4}\Bigl)^2 \; – \Bigl(\frac{5}{4}\Bigl)^2\biggl] \\
&= 2\biggl[\Bigl(x + \frac{1}{4} + \frac{5}{4}\Bigl)\Bigl(x + \frac{1}{4} \; – \frac{5}{4}\Bigl)\biggl] \\
&= 2\biggl[\Bigl(x + \frac{3}{2}\Bigl)(x \; – 1)\biggl] \end{align*}\)

3. การแยกตัวประกอบรวมทุกแบบ (โจทย์ระคน)

\(6a^2 + 9a\)
\(5a^4 \; – 10a^2\)
\(a^2 \; – 4\)
\(x^3 \; – x\)
\(x^2 \; – 1\)
\(a^3 + 2a^2\)
\(b^2 \; – 100\)
\(a^2 \; – b^2\)
\(x^2 \; – x\)
\(2x^2 + 6x^2\)
\(9c^2 \; – 16d^2\)
\(c^2 \; – 25\)
\(c^2 \; – 5c\)
\(m^2 \; – 4n^2\)
\(a^2b^2 \; – 9\)
\(2a^2 + a\)
\(4a^3 \; – a\)
\(a^3 \; – b^3\)
\(a^2 \; – 2ab + b^2\)
\(a^2 + ab\)
\(c^2 \; – 16\)
\(c^2 \; – 8c + 16\)
\(3c \; – 9\)
\(x^2 + 3x + 2\)
\(x^2 \; – 7x + 6\)
\(x^2 \; – 4x \; – 12\)
\(x^2 + 4x\)
\(x^3 \; – 9x\)
\(x^2 + 2x \; – 8\)
\(x^2 + x \; – 6\)
\(x^2 + 7x + 12\)
\(x^2 + 9x + 20\)
\(x^2 + 3x + 2\)
\(x^2 + 5x + 6\)
\(x^3 + 9x^2 + 20x\)
\(x^2 + 5x + 4\)
\(x^2 + 7x + 10\)
\(x^2 + 3x + 2\)
\(y^2 \; – 2y \; – 24\)
\(y^2 \; – y \; – 12\)
\(y^2 + 6y + 9\)
\(a^2 \; – 4a \; – 21\)
\(a^3 + 27\)
\(m^2 \; – m \; – 12\)
\(m^3 \; – 64\)
\(m^2 + m \; – 6\)
\(2y \; – y^2 \; – 6y^3\)
\(2 \; – 7y + 6y^2\)
\(4y + 12y^2 + 9y^3\)
\(4 \; – 9y^2\)
\(b^4 \; – 27b\)
\(4b^2 \; – 25\)
\(2b^2 + 5b\)
\(2b^2 \; – 11b + 5\)
\(3x^2 \; – 7x + 2\)

ดูเฉลยคำตอบ

\(6a^2 + 9a = 3a(2a + 3)\)
\(5a^4 \; – 10a^2 = 5a^2(a^2 \; – 2)\)
\(a^2 \; – 4 = (a \; – 2)(a + 2)\)
\(x^3 \; – x = x(x^2 \; – 1) = x(x \; – 1)(x + 1)\)
\(x^2 \; – 1 = (x \; – 1)(x + 1)\)
\(a^3 + 2a^2 = a^2(a + 2)\)
\(b^2 \; – 100 = b^2 \; – 10^2 = (b \; – 10)(b + 10)\)
\(a^2 \; – b^2 = (a \; – b)(a + b)\)
\(x^2 \; – x = x(x \; – 1)\)
\(2x^2 + 6x^2 = 8x^2\)
\(9c^2 \; – 16d^2 = (3c \; – 4d)(3c + 4d)\)
\(c^2 \; – 25 = (c \; – 5)(c + 5)\)
\(c^2 \; – 5c = c(c \; – 5)\)
\(m^2 \; – 4n^2 = (m \; – 2n)(m + 2n)\)
\(a^2b^2 \; – 9 = (ab \; – 3)(ab + 3)\)
\(2a^2 + a = a(2a + 1)\)
\(4a^3 \; – a = a(4a^2 \; – 1)\)
\(a^3 \; – b^3 = (a \; – b)(a^2 + ab + b^2)\)
\(a^2 \; – 2ab + b^2 = (a \; – b)^2\)
\(a^2 + ab = a(a + b)\)
\(c^2 \; – 16 = (c \; – 4)(c + 4)\)
\(c^2 \; – 8c + 16 = (c \; – 4)^2\)
\(3c \; – 9 = 3(c \; – 3)\)
\(x^2 + 3x + 2 = (x + 1)(x + 2)\)
\(x^2 \; – 7x + 6 = (x \; – 6)(x \; – 1)\)
\(x^2 \; – 4x \; – 12 = (x \; – 6)(x + 2)\)
\(x^2 + 4x = x(x + 4)\)
\(x^3 \; – 9x = x(x^2 \; – 9) = x(x \; – 3)(x + 3)\)
\(x^2 + 2x \; – 8 = (x \; – 2)(x + 4)\)
\(x^2 + x \; – 6 = (x \; – 2)(x + 3)\)
\(x^2 + 7x + 12 = (x + 3)(x + 4)\)
\(x^2 + 9x + 20 = (x + 4)(x + 5)\)
\(x^2 + 3x + 2 = (x + 2)(x + 1)\)
\(x^2 + 5x + 6 = (x + 3)(x + 2)\)
\(x^3 + 9x^2 + 20x = x(x^2 + 9x + 20) = (x)(x + 4)(x + 5)\)
\(x^2 + 5x + 4 = (x + 4)(x + 1)\)
\(x^2 + 7x + 10 = (x + 5)(x + 2)\)
\(x^2 + 3x + 2 = (x + 2)(x + 1)\)
\(y^2 \; – 2y \; – 24 = (y \; – 6)(y + 4)\)
\(y^2 \; – y \; – 12 = (y \; – 4)(y + 3)\)
\(y^2 + 6y + 9 = (y + 3)(y + 3) = (y + 3)^2\)
\(a^2 \; – 4a \; – 21 = (a \; – 7)(a + 3)\)
\(a^3 + 27 = a^3 + 3^3 = (a + 3)(a^2 \; – 3a + 9)\)
\(m^2 \; – m \; – 12 = (m \; – 4)(m + 3)\)
\(m^3 \; – 64 = m^3 \; – 4^3 = (m \; – 4)(m^2 + 4m + 16)\)
\(m^2 + m \; – 6 = (m \; – 2)(m + 3)\)
\(2y \; – y^2 \; – 6y^3 = (-y)(6y^2 + y \; – 2) = (-y)(2y \; – 1)(3y + 2)\)
\(2 \; – 7y + 6y^2 = 6y^2 \; – 7y + 2 = (3y \; – 2)(2y \; – 1)\)
\(4y + 12y^2 + 9y^3 = y(9y^2 + 12y + 4) = (y)(3y + 2)(3y + 2) = (y)(3y + 2)^2\)
\(4 \; – 9y^2 = (2 \; – 3y)(2 + 3y)\)
\(b^4 \; – 27b = b(b^3 \; – 27) = b(b \; – 3)(b^2 + 3b + 9)\)
\(4b^2 \; – 25 = (2b)^2 \; – 5^2 = (2b \; – 5)(2b + 5)\)
\(2b^2 + 5b = b(2b + 5)\)
\(2b^2 \; – 11b + 5 = (2b \; – 1)(b \; – 5)\)
\(3x^2 \; – 7x + 2 = (3x \; – 1)(x \; – 2)\)

1. จงแยกตัวประกอบของพหุนามต่อไปนี้

2. จงแยกตัวประกอบของพหุนามต่อไปนี้ โดยวิธีทำเป็นกำลังสองสมบูรณ์

3. การแยกตัวประกอบรวมทุกแบบ (โจทย์ระคน)

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