Question

phân tích nhân tử

a, \(x^2\) +4x+3

b, 2\(x^2\) +3x-5

c, 16x-5\(x^2\)-3

Answer 1

Lời giải:
a)

\(x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)\)

\(=(x+3)(x+1)\)

b)

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

\(=2x(x-1)+5(x-1)=(x-1)(2x+5)\)

c)

\(16x-5x^2-3=(15x-5x^2)+(x-3)\)

\(5x(3-x)-(3-x)=(5x-1)(1-3x)\)

Answer 2

\(a,x^2+4x+3=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+3\right)\left(x+1\right)\)

\(b,2x^2+3x-5\)

\(=2x^2-2x+5x-5\)

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

\(c,16x-5x^2-3\)

\(=-5x^2+15x+x-3\)

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