1. จงแยกตัวประกอบของพหุนามต่อไปนี้
1) \(\small\mathsf{9(y^2 \; – \, 20y + 100) \; – \, 441y^2}\)
วิธีทำ
\(\small\mathsf{9(y^2 \; – \, 20y + 100) \; – \, 441y^2}\)
= \(\small\mathsf{9[(y^2 \; – \, 20y + 100) \; – \, 49y^2]}\)
= \(\small\mathsf{9[(y^2 \; – \, 2(y)(10) + 10^2) \; – \, {7y}^2]}\)
= \(\small\mathsf{9[{(y \; – \, 10)}^2 \; – \, {7y}^2]}\)
= 9[(y – 10 + 7y)(y – 10 – 7y)]
= 9[(8y – 10)(-6y – 10)]
= 9(2)(4y – 5)(-2)(3y + 5)
= (-36)(4y – 5)(3y + 5)
= \(\small\mathsf{9[{(y \; – \, 10)}^2 \; – \, {7y}^2]}\)
= 9[(y – 10 + 7y)(y – 10 – 7y)]
= 9[(8y – 10)(-6y – 10)]
= 9(2)(4y – 5)(-2)(3y + 5)
= (-36)(4y – 5)(3y + 5)
ตอบ (-36)(4y – 5)(3y + 5)
2) \(\small\mathsf{(4t^2 \; – \, 12t + 9) \; – \, {(5t + 1)}^2}\)
วิธีทำ
\(\small\mathsf{(4t^2 \; – \, 12t + 9) \; – \, {(5t + 1)}^2}\)
= \(\small\mathsf{[{(2t)}^2 \; – \, 2(2t)(3) + 3^2] \; – \, {(5t + 1)}^2}\)
= \(\small\mathsf{{(2t \; – \, 3)}^2 \; – \, {(5t + 1)}^2}\)
= [(2t – 3) + (5t + 1)][(2t – 3) – (5t + 1)]
= (2t – 3 + 5t + 1)(2t – 3 – 5t – 1)
= (7t – 2)(-3t – 4)
= (7t – 2)(-1)(3t + 4)
= (-1)(7t – 2)(3t + 4)
= [(2t – 3) + (5t + 1)][(2t – 3) – (5t + 1)]
= (2t – 3 + 5t + 1)(2t – 3 – 5t – 1)
= (7t – 2)(-3t – 4)
= (7t – 2)(-1)(3t + 4)
= (-1)(7t – 2)(3t + 4)
ตอบ (-1)(7t – 2)(3t + 4)
3) (4\(\small\mathsf{m^2}\) – 36m + 81) – (16\(\small\mathsf{m^2}\) + 56m + 49)
วิธีทำ
(4\(\small\mathsf{m^2}\) – 36m + 81) – (16\(\small\mathsf{m^2}\) + 56m + 49)
= [\(\small\mathsf{{(2m)}^2}\) – 2(2m)(9) + \(\small\mathsf{9^2}\)] – [\(\small\mathsf{{(4m)}^2}\) + 2(4m)(7) + \(\small\mathsf{7^2}\)]
= \(\small\mathsf{{(2m \; – \, 9)}^2 \; – \, {(4m + 7)}^2}\)
= [(2m – 9) + (4m + 7)][(2m – 9) – (4m + 7)]
= (2m – 9 + 4m + 7)(2m – 9 – 4m – 7)
= (6m – 2)(-2m – 16)
= (2)(3m – 1)(-2)(m + 8)
= (-4)(3m – 1)(m + 8)
= [(2m – 9) + (4m + 7)][(2m – 9) – (4m + 7)]
= (2m – 9 + 4m + 7)(2m – 9 – 4m – 7)
= (6m – 2)(-2m – 16)
= (2)(3m – 1)(-2)(m + 8)
= (-4)(3m – 1)(m + 8)
ตอบ (-4)(3m – 1)(m + 8)
4) \(\small\mathsf{{(n^2 + 3n + 1)}^2 \; – \, 1}\)
วิธีทำ
\(\small\mathsf{{(n^2 + 3n + 1)}^2 \; – \, 1}\)
= \(\small\mathsf{{(n^2 + 3n + 1)}^2 \; – \, 1^2}\)
= \(\small\mathsf{[(n^2 + 3n + 1) + 1][(n^2 + 3n + 1) \; – \, 1]}\)
= \(\small\mathsf{(n^2 + 3n + 2)(n^2 + 3n)}\)
= (n + 1)(n + 2)(n)(n + 3)
= n(n + 1)(n + 2)(n + 3)
= \(\small\mathsf{(n^2 + 3n + 2)(n^2 + 3n)}\)
= (n + 1)(n + 2)(n)(n + 3)
= n(n + 1)(n + 2)(n + 3)
ตอบ n(n + 1)(n + 2)(n + 3)
2. ถ้า \(\small\mathsf{7x^3 + 14x^2 = A(x + 2)}\) และ \(\small\mathsf{B(y + 5) = y^2 + 2y \; – \, 15}\) โดยที่ A และ B เป็นพหุนามแล้ว A + 2B มีค่าเท่าใด
วิธีทำ
หาค่า A จาก
\(\small\mathsf{7x^3 + 14x^2}\)
\(\small\mathsf{7x^2(x + 2)}\)
\(\mathsf{\frac{7x^2(x + 2)}{x + 2}}\)
\(\small\mathsf{7x^2(x + 2)}\)
\(\mathsf{\frac{7x^2(x + 2)}{x + 2}}\)
= A(x + 2)
= A(x + 2)
= A
= A(x + 2)
= A
\(\small\mathsf{7x^2}\) = A
จะได้ A = \(\small\mathsf{7x^2}\)
หาค่า B จาก
หาค่า B จาก
B(y + 5)
B(y + 5)
\(\mathsf{\frac{B(y + 5)}{y + 5}}\)
B(y + 5)
\(\mathsf{\frac{B(y + 5)}{y + 5}}\)
= \(\small\mathsf{y^2}\) + 2y – 15
= (y – 3)(y + 5)
= y – 3
= (y – 3)(y + 5)
= y – 3
B = y – 3
จะได้ B = y – 3
ดังนั้น A + 2B = \(\small\mathsf{7x^2}\) + 2(y – 3) = \(\small\mathsf{7x^2}\) + 2y – 6
ดังนั้น A + 2B = \(\small\mathsf{7x^2}\) + 2(y – 3) = \(\small\mathsf{7x^2}\) + 2y – 6
ตอบ \(\small\mathsf{7x^2}\) + 2y – 6
3. กำหนดให้ a + b = 2 และ \(\small\mathsf{b^2 \, – \, a^2}\) = 10 จงหาค่าของ a – b
วิธีทำ
\(\small\mathsf{b^2 \, – \, a^2}\)
= 10
(b + a)(b – a)
(a + b)(b – a)
2(b – a)
b – a
b – a
(-1)(-b + a)
-b + a
a – b
(a + b)(b – a)
2(b – a)
b – a
b – a
(-1)(-b + a)
-b + a
a – b
= 10
= 10
= 10
= \(\small\mathsf{\frac{10}{2}}\)
= 5
= 5
= \(\small\mathsf{\frac{5}{(-1)}}\)
= -5
= 10
= 10
= \(\small\mathsf{\frac{10}{2}}\)
= 5
= 5
= \(\small\mathsf{\frac{5}{(-1)}}\)
= -5
ตอบ -5
4. กำหนดให้ D = \(\mathsf{\frac{2,558^2 \, – \, 2,550^2 \, – \, 8^2}{2,550}}\) จงหาค่าของ D
วิธีทำ
D
= \(\mathsf{\frac{2,558^2 \, – \, 2,550^2 \, – \, 8^2}{2,550}}\)
= \(\mathsf{\frac{(2,558^2 \, – \, 2,550^2) \, – \, 8^2}{2,550}}\)
= \(\mathsf{\frac{(2,558 \, + \, 2,550)(2,558 \, – \, 2,550) \, – \, 8^2}{2,550}}\)
= \(\mathsf{\frac{(5,108)(8) \, – \, 8^2}{2,550}}\)
= \(\mathsf{\frac{8(5,108 \, – \, 8)}{2,550}}\)
= \(\mathsf{\frac{8(5,100)}{2,550}}\)
= 8 x 2
= 16
ตอบ 16