\(=14x^2-2x+35x-5\)

\(=2x\left(7x-1\right)+5\left(7x-1\right)\)

\(\left(7x-1\right)\left(2x+5\right)\)

Phân tích đa thức thành nhân tử: 14x^2 + 33x - 5

=14x2−2x+35x−5

=2x(7x−1)+5(7x−1)

(7x−1)(2x+5)

Phân tích đa thức thành nhân tử: 14x^2 + 33x - 5 là :

         \(14x^2-2x+35x-5\)

          \(2x\left(7x-1\right)+5\left(7x-1\right)\)

           \(\Rightarrow\left(7x-1\right)\left(2x+5\right).\)

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).     b) (3x – y – 1)(x – 7y – 1).

a) \(x^3+6x^2+3x-10\)

\(=x^3-x^2+7x^2-7x+10x-10\)

\(=x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+7x+10\right)\)

\(=\left(x-1\right)\left(x^2+2x+5x+10\right)\)

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

b)  \(x^3+3x^2-33x-35\)

\(=x^3-5x^2+8x^2-40x+7x-35\)

\(=x^2\left(x-5\right)+8x\left(x-5\right)+7\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+8x+7\right)\)

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

\(=\left(x-5\right)\left(x+1\right)\left(x+7\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

b) 3x⁴-3x³+9x³-9x²-24x²+24x-48x+48

=3x³(x-1)+9x²(x-1)-24x(x-1)-48(x-1)

=(x-1)(3x³+9x²-24x-48)

=3(x-1)(x³+3x²-8x-16)

\(=6x^2+3x+10x+5=3x\left(2x+1\right)+5\left(2x+1\right)=\left(3x+5\right)\left(2x+1\right)\)

6x2+3x+10x+5=3x(2x+1)+5(2x+1)=(3x+5)(2x+1)