3. จงแยกตัวประกอบของพหุนามต่อไปนี้
วิธีคิด
สูตรการแยกตัวประกอบของพหุนามดีกรีสองที่เป็นกำลังสองสมบูรณ์
เมื่อ A แทนพจน์หน้า และ B แทนพจน์หลัง
\(\small\mathsf{A^2 + 2AB + B^2 = {(A + B)}^2}\)
\(\small\mathsf{A^2 \, – \, 2AB + B^2 = {(A \, – \, B)}^2}\)
\(\small\mathsf{A^2 \, – \, 2AB + B^2 = {(A \, – \, B)}^2}\)
หรือสามารถเขียนสูตรให้จำได้ง่ายขึ้น ดังนี้
\(\small\mathsf{หน้า^2 + 2(หน้า)(หลัง) + หลัง^2 = (หน้า + หลัง)^2}\)
\(\small\mathsf{หน้า^2 \, – \, 2(หน้า)(หลัง) + หลัง^2 = (หน้า \, – \, หลัง)^2}\)
\(\small\mathsf{หน้า^2 \, – \, 2(หน้า)(หลัง) + หลัง^2 = (หน้า \, – \, หลัง)^2}\)
1) \(\small\mathsf{{(x \, – \, 2)}^2 + 12(x \, – \, 2) + 36}\)
วิธีทำ
\(\small\mathsf{{(x \, – \, 2)}^2 + 12(x \, – \, 2) + 36}\)
= \(\small\mathsf{{(x \, – \, 2)}^2 + 2(x \, – \, 2)(6) + 6^2}\)
= \(\small\mathsf{{(x \, – \, 2 + 6)}^2}\)
= \(\small\mathsf{{(x + 4)}^2}\)
= \(\small\mathsf{{(x + 4)}^2}\)
ตอบ \(\small\mathsf{{(x + 4)}^2}\)
2) \(\small\mathsf{{(2x + 1)}^2 + 20(2x + 1) + 100}\)
วิธีทำ
\(\small\mathsf{{(2x + 1)}^2 + 20(2x + 1) + 100}\)
= \(\small\mathsf{{(2x + 1)}^2 + 2(2x + 1)(10) + {10}^2}\)
= \(\small\mathsf{{(2x + 1 + 10)}^2}\)
= \(\small\mathsf{{(2x + 11)}^2}\)
= \(\small\mathsf{{(2x + 11)}^2}\)
ตอบ \(\small\mathsf{{(2x + 11)}^2}\)
3) \(\small\mathsf{{(x + 3)}^2 \, – \, 16(x + 3) + 64}\)
วิธีทำ
\(\small\mathsf{{(x + 3)}^2 \, – \, 16(x + 3) + 64}\)
= \(\small\mathsf{{(x + 3)}^2 \, – \, 2(x + 3)(8) + 8^2}\)
= \(\small\mathsf{{(x + 3 \, – \, 8)}^2}\)
= \(\small\mathsf{{(x \, – \, 5)}^2}\)
= \(\small\mathsf{{(x \, – \, 5)}^2}\)
ตอบ \(\small\mathsf{{(x \, – \, 5)}^2}\)
4) \(\small\mathsf{{(4x \, – \, 5)}^2 \, – \, 26(4x \, – \, 5) + 169}\)
วิธีทำ
\(\small\mathsf{{(4x \, – \, 5)}^2 \, – \, 26(4x \, – \, 5) + 169}\)
= \(\small\mathsf{{(4x \, – \, 5)}^2 \, – \, 2(4x \, – \, 5)(13) + {13}^2}\)
= \(\small\mathsf{{(4x \, – \, 5 \, – \, 13)}^2}\)
= \(\small\mathsf{{(4x \, – \, 18)}^2}\)
= \(\small\mathsf{{[2(2x \, – \, 9)]}^2}\)
= \(\small\mathsf{{2^2(2x \, – \, 9)}^2}\)
= \(\small\mathsf{{4(2x \, – \, 9)}^2}\)
= \(\small\mathsf{{(4x \, – \, 18)}^2}\)
= \(\small\mathsf{{[2(2x \, – \, 9)]}^2}\)
= \(\small\mathsf{{2^2(2x \, – \, 9)}^2}\)
= \(\small\mathsf{{4(2x \, – \, 9)}^2}\)
ตอบ \(\small\mathsf{{4(2x \, – \, 9)}^2}\)
5) \(\small\mathsf{36{(x + 6)}^2 + 108(x + 6) + 81}\)
วิธีทำ
\(\small\mathsf{36{(x + 6)}^2 + 108(x + 6) + 81}\)
= \(\small\mathsf{9[4{(x + 6)}^2 + 12(x + 6) + 9]}\)
= \(\small\mathsf{9[2^2{(x + 6)}^2 + (2)(2)(3)(x + 6) + 3^2]}\)
= \(\small\mathsf{9[{(2(x + 6))}^2 + 2(2(x + 6))(3) + 3^2]}\)
= \(\small\mathsf{9{[2(x + 6) + 3]}^2}\)
= \(\small\mathsf{9{(2x + 12 + 3)}^2}\)
= \(\small\mathsf{9{(2x + 15)}^2}\)
= \(\small\mathsf{9[{(2(x + 6))}^2 + 2(2(x + 6))(3) + 3^2]}\)
= \(\small\mathsf{9{[2(x + 6) + 3]}^2}\)
= \(\small\mathsf{9{(2x + 12 + 3)}^2}\)
= \(\small\mathsf{9{(2x + 15)}^2}\)
ตอบ \(\small\mathsf{9{(2x + 15)}^2}\)
6) \(\small\mathsf{9{(x \, – \, 1)}^2 \, – \, 30(x \, – \, 1) + 25}\)
วิธีทำ
\(\small\mathsf{9{(x \, – \, 1)}^2 \, – \, 30(x \, – \, 1) + 25}\)
= \(\small\mathsf{3^2{(x \, – \, 1)}^2 \, – \, (2)(3)(5)(x \, – \, 1) + 5^2}\)
= \(\small\mathsf{{[3(x \, – \, 1)]}^2 \, – \, (2)[3(x \, – \, 1)](5) + 5^2}\)
= \(\small\mathsf{{[3(x \, – \, 1) \, – \, 5]}^2}\)
= \(\small\mathsf{{(3x \, – \, 3 \, – \, 5)}^2}\)
= \(\small\mathsf{{(3x \, – \, 8)}^2}\)
= \(\small\mathsf{{[3(x \, – \, 1) \, – \, 5]}^2}\)
= \(\small\mathsf{{(3x \, – \, 3 \, – \, 5)}^2}\)
= \(\small\mathsf{{(3x \, – \, 8)}^2}\)
ตอบ \(\small\mathsf{{(3x \, – \, 8)}^2}\)
7) \(\small\mathsf{16x^2 + 8x(x + 1) + {(x + 1)}^2}\)
วิธีทำ
\(\small\mathsf{16x^2 + 8x(x + 1) + {(x + 1)}^2}\)
= \(\small\mathsf{{(4x)}^2 + 2(4x)(x + 1) + {(x + 1)}^2}\)
= \(\small\mathsf{{(4x + x + 1)}^2}\)
= \(\small\mathsf{{(5x + 1)}^2}\)
= \(\small\mathsf{{(5x + 1)}^2}\)
ตอบ \(\small\mathsf{{(5x + 1)}^2}\)
8) \(\small\mathsf{{(x \, – \, 3)}^2 \, – \, 12x(x \, – \, 3) + 36x^2}\)
วิธีทำ
\(\small\mathsf{{(x \, – \, 3)}^2 \, – \, 12x(x \, – \, 3) + 36x^2}\)
= \(\small\mathsf{{(x \, – \, 3)}^2 \, – \, 2(x \, – \, 3)(6x) + {(6x)}^2}\)
= \(\small\mathsf{{(x \, – \, 3 \, – \, 6x)}^2}\)
= \(\small\mathsf{{(-5x \, – \, 3)}^2}\)
= \(\small\mathsf{{[(-1)(5x + 3)]}^2}\)
= \(\small\mathsf{{(-1)}^2{(5x + 3)}^2}\)
= \(\small\mathsf{{(5x + 3)}^2}\)
= \(\small\mathsf{{(-5x \, – \, 3)}^2}\)
= \(\small\mathsf{{[(-1)(5x + 3)]}^2}\)
= \(\small\mathsf{{(-1)}^2{(5x + 3)}^2}\)
= \(\small\mathsf{{(5x + 3)}^2}\)
ตอบ \(\small\mathsf{{(5x + 3)}^2}\)
9) \(\small\mathsf{49x^2 + 14(x^2 \, – \, x) + {(x \, – \, 1)}^2}\)
วิธีทำ
\(\small\mathsf{49x^2 + 14(x^2 \, – \, x) + {(x \, – \, 1)}^2}\)
= \(\small\mathsf{{(7x)}^2 + 2(7)[x(x \, – \, 1)] + {(x \, – \, 1)}^2}\)
= \(\small\mathsf{{(7x)}^2 + 2(7x)(x \, – \, 1) + {(x \, – \, 1)}^2}\)
= \(\small\mathsf{{(7x + x \, – \, 1)}^2}\)
= \(\small\mathsf{{(8x \, – \, 1)}^2}\)
= \(\small\mathsf{{(7x + x \, – \, 1)}^2}\)
= \(\small\mathsf{{(8x \, – \, 1)}^2}\)
ตอบ \(\small\mathsf{{(8x \, – \, 1)}^2}\)
10) \(\small\mathsf{{(x + 2)}^2 \, – \, 18(x^2 + 2x) + 81x^2}\)
วิธีทำ
\(\small\mathsf{{(x + 2)}^2 \, – \, 18(x^2 + 2x) + 81x^2}\)
= \(\small\mathsf{{(x + 2)}^2 \, – \, 2(9)[x(x + 2)] + {(9x)}^2}\)
= \(\small\mathsf{{(x + 2)}^2 \, – \, 2(x + 2)(9x) + {(9x)}^2}\)
= \(\small\mathsf{{(x + 2 \, – \, 9x)}^2}\)
= \(\small\mathsf{{(2 \, – \, 8x)}^2}\)
= \(\small\mathsf{{[2(1 \, – \, 4x)]}^2}\)
= \(\small\mathsf{2^2{(1 \, – \, 4x)}^2}\)
= \(\small\mathsf{4{(1 \, – \, 4x)}^2}\)
= \(\small\mathsf{{(x + 2 \, – \, 9x)}^2}\)
= \(\small\mathsf{{(2 \, – \, 8x)}^2}\)
= \(\small\mathsf{{[2(1 \, – \, 4x)]}^2}\)
= \(\small\mathsf{2^2{(1 \, – \, 4x)}^2}\)
= \(\small\mathsf{4{(1 \, – \, 4x)}^2}\)
ตอบ \(\small\mathsf{4{(1 \, – \, 4x)}^2}\)