$$(x-2)(9\times 2-8x)=(x-2)(16x-16)$$

Jaylyn Horne
2022-10-26
Answered

Find all Real solutions to this equation:

$$(x-2)(9\times 2-8x)=(x-2)(16x-16)$$

$$(x-2)(9\times 2-8x)=(x-2)(16x-16)$$

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toliwask

Answered 2022-10-27
Author has **15** answers

$$(x-2)(9\times 2-8x)=(x-2)(16x-16)\phantom{\rule{0ex}{0ex}}9{x}^{3}-26{x}^{2}+16x=16{x}^{2}-48x+32$$

Subtract $$16{x}^{2}-48x+32$$ from both sides

$$9{x}^{3}-26{x}^{2}+16x-16{x}^{2}+48x-32=0\phantom{\rule{0ex}{0ex}}9{x}^{3}-42{x}^{2}+64x-32=0$$

One factor of this polynomial is x-2

Divide the polynomial $$9{x}^{3}-42{x}^{2}+64x-32$$ by x-2. So we get:

$$(9{x}^{3}-42{x}^{2}+64x-32)/(x-2)=9{x}^{2}-24x+16$$

Our polynomial becomes:

$$(9{x}^{2}-24x+16)=(3x-4{)}^{2}\phantom{\rule{0ex}{0ex}}(x-2)(3x-4{)}^{2}=0$$

see factors equal to 0

x-2=0 or 3x-4=0

x=2 or x=4/3

Real solutions are x=2, 4/3

Subtract $$16{x}^{2}-48x+32$$ from both sides

$$9{x}^{3}-26{x}^{2}+16x-16{x}^{2}+48x-32=0\phantom{\rule{0ex}{0ex}}9{x}^{3}-42{x}^{2}+64x-32=0$$

One factor of this polynomial is x-2

Divide the polynomial $$9{x}^{3}-42{x}^{2}+64x-32$$ by x-2. So we get:

$$(9{x}^{3}-42{x}^{2}+64x-32)/(x-2)=9{x}^{2}-24x+16$$

Our polynomial becomes:

$$(9{x}^{2}-24x+16)=(3x-4{)}^{2}\phantom{\rule{0ex}{0ex}}(x-2)(3x-4{)}^{2}=0$$

see factors equal to 0

x-2=0 or 3x-4=0

x=2 or x=4/3

Real solutions are x=2, 4/3

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