Question

Phân tích các đa thức sau thành nhân tử:

A₄=(x²-3x+5)2 + 7x(x²-3x +5) + 12x²

A₅=2(x²+5x-2)²-7(x²+5x-2)(x³+3)+5(x²+3)²

A₆=4(x²-15x+50)(x²-18x+72)-3x²

A₇=x⁴-7x³+12x²-x-3

Answer 1

a) \(A_4=\left(x^2-3x+5\right)^2+7x\cdot\left(x^2-3x+5\right)+12x^2\)

\(=\left(x^2-3x+5\right)^2+4x\cdot\left(x^2-3x+5\right)+3x\left(x^2-3x+5\right)+12x^2\)

\(=\left(x^2-3x+5\right)\left(x^2-3x+5+4x\right)+3x\left(x^2-3x+5+4x\right)\)

\(=\left[\left(x^2-3x+5\right)+3x\right]\cdot\left(x^2-3x+5+4x\right)\)

\(=\left(x^2-3x+5+3x\right)\left(x^2+x+5\right)\)

\(=\left(x^2+5\right)\left(x^2+x+5\right)\)

Answer 2

\(A_5=2\left(x^2+5x-2\right)^2-7\left(x^2+5x-2\right)\left(x^3+3\right)+5\left(x^2+3\right)^2\)

Đặt \(x^2+5x-2=a;x^3+3=b\),Ta có:

\(2a^2-7ab+5b^2=2a^2-5ab-2ab+5b^2=a\left(2a-5b\right)-b\left(2a-5b\right)=\left(2a+5b\right)\left(a-b\right)\)

Thay \(x^2+5x-2=a;x^3+3=b\),ta có:

.......................

bn làm nốt nhé

Answer 3

\(4\left(x^2-15x+50\right)\left(x^2-18+72\right)-3x^2\)

\(=4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\)

\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+10\right)\left(x+6\right)\right]-3x^2\)

\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

Đặt \(x^2+16x+60=a\), ta có:

\(4\left(a+x\right)\left(a\right)-3x^2\)

\(=4a^2+4ax-3x^2\)

\(=4a^2-2ax+6ax-3x^2=2a\left(2a-x\right)+3x\left(2a-x\right)\)

\(=\left(2a-x\right)\left(2a+3x\right)\)

Thay a vào ta có: \(\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\)

\(=\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

Answer 4

\(A_7=x^4-7x^3+12x^2-x-3\)

\(=\left(x^4-2x^3-x^2\right)+\left(5x^3-10x^2-5x\right)+\left(3x^2-6x-3\right)\)

\(=x^2\left(x^2-2x-1\right)+5x\left(x^2-2x-1\right)+3\left(x^2+2x-1\right)\)

\(=\left(x^2-2x-1\right)\left(x^2+5x+3\right)\)