Giải SBT Toán 8 CTST Bài 4. Phân tích đa thức thành nhân tử có đáp án

a) 3x² + 6xy = 3x.x + 3x.2y = 3x(x + 2y).

b) 5(y – 3) – x(3 – y)

= 5(y – 3) + x(y ‒ 3)

= (y ‒ 3)(5 + x).

c) 2x³ – 6x² = 2x².x‒ 2x².3 = 2x²(x ‒ 3).

d) x⁴y² + xy³ = xy².x³ + xy².y = xy²(x³ + y).

e) xy – 2xyz + x²y

= xy ‒ xy.2z + xy.x

= xy(1 ‒ 2z + x).

g) (x + y)³ – x(x + y)²

= (x + y)².(x + y) – x(x + y)²

= (x + y)² (x + y ‒ x)

= y(x + y)².

a) 100 – x² = 10²– x² = (10 ‒ x)(10 + x).

b) 4x² – y² = (2x)²‒ y² = (2x ‒ y)(2x + y).

c) x+y2−14y2;

=x+y2−12y2

=x+y−12yx+y+12y

=x+12yx+32y.

d) (x – y)² – (y – z)²

= (x ‒ y + y ‒ z)(x ‒ y ‒ y + z)

= (x ‒ z)(x ‒ 2y + z).

e) x² – (1 + 2x)²

= (x + 1 + 2x)(x ‒ 1 ‒ 2x)

= (3x + 1)(‒x ‒ 1).

g) x⁴ – 16 = (x²)²‒ 4²

= (x² ‒ 4)(x² + 4)

= (x² ‒ 2²)(x² + 4)

= (x + 2)(x ‒ 2)(x² + 4).

a) a² + 12a + 36

= a² + 2.a.6 + 6²

= (a + 6)².

b) –9 + 6a – a²

= ‒(a² ‒ 6a + 9)

= ‒(a² ‒ 2.3.a + 3²)

= ‒(a ‒ 3)².

c) 2a² + 8b² – 8ab

= 2(a² + 4b² ‒ 4ab)

= 2[a² ‒ 2.a.2b + (2b)²]

= 2(a ‒ 2b)².

d) 16a² + 8ab² + b⁴

= (4a)² + 2.4a.b² + (b²)²

= (4a + b²)².

a) x³ – 1 000

= x³‒ 10³

= (x ‒ 10)(x² + 10x + 10²)

= (x ‒ 10)(x² + 10x + 100).

b) 8x³ + (x – y)³

= (2x)³ + (x – y)³

= (2x + x ‒ y)[(2x)² ‒ 2x(x ‒ y) + (x ‒ y)²]

= (3x ‒ y)(4x² ‒ 2x² + 2xy + x² ‒ 2xy + y²)

= (3x ‒ y)[(4x² ‒ 2x² + x²) + (2xy ‒ 2xy) + y²]

= (3x ‒ y)(3x² + y²).

c) (x – 1)³ – 27

= (x – 1)³ – 3³

= (x ‒ 1 ‒ 3)[(x ‒ 1)² + (x ‒ 1).3 + 3²]

= (x ‒ 4)(x² ‒ 2x + 1 + 3x ‒ 3 + 9)

= (x ‒ 4)[x² + (‒2x + 3x) + 1 ‒ 3 + 9]

= (x ‒ 4)(x² + x +7).

d) x⁶ + y⁹

= (x²)³ + (y³)³

= (x² + y³)[(x²)² ‒ x².y³ + (y³)²]

= (x² + y³)(x⁴ ‒ x²y³ + y⁶).

a) x + 2x(x – y) – y

= (x ‒ y) + 2x(x ‒y)

= (x ‒ y)(1 + 2x).

b) Cách 1:

x² + xy – 3x – 3y

= (x² + xy) – (3x + 3y)

= x(x + y) – 3(x + y)

= (x + y)(x – 3).

Cách 2:

x² + xy – 3x – 3y

= (x² ‒ 3x) + (xy ‒ 3y)

= x(x ‒ 3) + y(x ‒ 3)

= (x ‒ 3)(x + y).

c) Cách 1:

xy – 5y + 4x – 20

= (xy – 5y) + (4x – 20)

= y(x – 5) + 4(x – 5)

= (x – 5)(y + 4).

Cách 2:

xy – 5y + 4x – 20

= (xy + 4x) ‒ (5y + 20)

= x(y + 4) ‒ 5(y + 4)

= (y + 4)(x ‒ 5).

d) Cách 1:

5xy – 25x² + 50x – 10y

= (5xy – 25x²) + (50x – 10y)

= 5x(y ‒ 5x) + 10(5x ‒ y)

= 5x(y ‒ 5x) ‒ 10(y ‒ 5x)

= 5(y ‒ 5x)(x ‒ 2).

Cách 2:

5xy – 25x² + 50x – 10y

= (5xy – 10y) – (25x² – 50x)

= 5y(x – 2) – 25x(x – 2)

= 5(x – 2)(y – 5x).

a) P = 7(a − 4) – b(4 – a) = 7(a − 4) + b(a ‒ 4) = (a ‒ 4)(7 + b).

Với a = 17 và b = 3 ta có:

P = (17 ‒ 4)(7 + 3) = 13.10 = 130.

b) Q = a² + 2ab – 5a – 10b = (a² + 2ab) – (5a + 10b)

= a(a + 2b) ‒ 5(a + 2b)= (a + 2b)(a ‒ 5).

Với a = 1,2 và b = 4,4 ta có:

Q = (1,2 + 2.4,4).(1,2 ‒ 5) = (1,2 + 8,8).(‒3,8) = 10. (‒3,8) = 38.

a) 4a² – 4b² – a – b

= (4a² – 4b²) – (a + b)

= 4(a² ‒ b²)– (a + b)

= 4(a ‒ b)(a + b)– (a + b)

= (a + b)(4a ‒ 4b ‒ 1).

b) 9a² – 4b² + 4b – 1

= 9a² – (4b² ‒ 4b + 1)

= (3a)² ‒ [(2b)² ‒ 2.2b + 1²]

= (3a)² ‒ (2b ‒ 1)²

= (3a + 2b ‒ 1)(3a ‒ 2b + 1)

c) 4x³ – y³ + 4x²y – xy²

= (4x³+ 4x²y) – (y³+ xy²)

= 4x²(x + y) ‒ y²(y + x)

= (x + y)(4x² ‒ y²)

= (x + y)[(2x)² ‒ y²]

= (x + y)(2x + y)(2x ‒ y).

d) a³ – b³ + 4ab + 4a² + 4b²

= (a³ – b³)+ (4a² + 4ab + 4b²)

= (a ‒ b)(a² + ab + b²) + 4.(a² + ab + b²)

= (a² + ab + b²)(a – b + 4).

a) 4x³ – 36x

= 4x(x²‒ 9)

= 4x(x² ‒ 3²)

= 4(x ‒ 3)(x + 3).

b) 4xy² – 4x²y – y³

= –y(–4xy + 4x² + y²)

= ‒y(4x²‒ 4xy + y²)

= ‒y[(2x)² ‒ 2.2x.y + y²]

= ‒y(2x – y)².

c) x⁶ – 64

= (x³)² ‒ 8²

= (x³ + 8)(x³ ‒ 8)

= (x³ + 2³)(x³ ‒ 2³)

= (x + 2)(x² ‒ 2x + 4)(x ‒ 2)(x² + 2x + 4 ).