Question

1) (1-x)(5x+3)=(3x-7)(x-1)

2) (x-2)(x+1)=x²-4

3) 2x³+3x²-32x=48

4) x²+2x-15=0

5) 2x(2x-3)=(3-2x)(2-5x)

6) x³-5x²+6x=0

7) (x²-5)(x+3)=0

8) (x+7)(3x-1)=49-x²

Answer 1

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

Answer 2

\(\left(x-2\right)\left(x+1\right)=x^2-4\)

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)

Answer 3

\(2x^3+3x^2-32x=48\)

\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)

\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)

\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)

Answer 4

\(x^2+2x-15=0\)

\(< =>x^2-3x+5x-15=0\)

\(< =>x\left(x-3\right)+5\left(x-3\right)=0\)

\(< =>\left(x+5\right)\left(x-3\right)=0\)

\(< =>\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

Answer 5

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

\(< =>2x\left(2x-3\right)-\left(3-2x\right)\left(2-5x\right)=0\)

\(< =>2x\left(2x-3\right)+\left(2x-3\right)\left(2-5x\right)=0\)

\(< =>\left(2x-3\right)\left(2x+2-5x\right)=0\)

\(< =>\left(2x-3\right)\left(2-3x\right)=0\)

\(< =>\orbr{\begin{cases}2x-3=0\\2-3x=0\end{cases}< =>\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{2}{3}\end{cases}}}\)

Answer 6

\(x^3-5x^2+6x=0\)

\(< =>x\left(x^2-5x+6\right)=0\)

\(< =>x\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)

\(< =>x\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)

\(< =>x\left(x-2\right)\left(x-3\right)=0\)

\(< =>\hept{\begin{cases}x=0\\x=2\\x=3\end{cases}}\)

Answer 7

\(\left(x^2-5\right)\left(x+3\right)=0\)

\(< =>\left(x^2-\sqrt{5}^2\right)\left(x+3\right)=0\)

\(< =>\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\left(x+3\right)=0\)

\(< =>\hept{\begin{cases}x-\sqrt{5}=0\\x+\sqrt{5}=0\\x+3=0\end{cases}}\)

\(< =>\hept{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\\x=-3\end{cases}}\)

Answer 8

\(\left(x+7\right)\left(3x-1\right)=49-x^2\)

\(< =>\left(x+7\right)\left(3x-1\right)+x^2-7^2=0\)

\(< =>\left(x+7\right)\left(3x-1\right)+\left(x-7\right)\left(x+7\right)=0\)

\(< =>\left(x+7\right)\left(3x-1+x-7\right)=0\)

\(< =>\left(x+7\right)\left(4x-8\right)=0\)

\(< =>4\left(x+7\right)\left(x-2\right)=0\)

\(< =>\orbr{\begin{cases}x+7=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-7\\x=2\end{cases}}}\)