(*x *+ 3)(*x *− 1) = − *x *+ 1 Solve for x.

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`(x+3)(x-1) = -x+1`

`(x+3)(x-1) = -(x-1)`

See that (x-1) term in both sides cancel out.

`(x+3) = -1`

`x = -1-3`

`x = -4`

*So the answer is x = -4.*

**It is quadratc equation so it has two roots**

`(x+3)(x-1)=-x+1`

`(x+3)(x-1)=-(x-1)`

`(x+3)(x-1)+(x-1)=0`

`(x-1)(x+3+1)=0`

`(x-1)(x+4)=0`

`either `

`x+4=0`

`x=-4`

`or`

`x-1=0`

`x=1`

**Also we can verify it wth help of graph**

**Let**

f(x)=(x+3)(x-1)+(x-1) The zeros of f(x) is root of equation f(x)=0

f(x) eets x-axis at x=-4 and x=1

(x+3)(x-1)=-X+1

we expand the brackets

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

x^2-x+3x-3=-x+1

x^2+3x-4=0

(x^2-x)+(4x-4)=0

x((x-1)+4(x-1)=0

(x+4)(x-1)=0

x=-4 or 1

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