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

$$= ( x - 3 ) ( x - 5 ) \cdot 2 x ( x - 3 )$$

Evaluate

$$2x\left(x-5\right)\left(x-3\right)^{2}$$

Solution Steps

Multiply x-3 and x-3 to get \left(x-3\right)^{2}.

$$\left(x-3\right)^{2}\left(x-5\right)\times 2x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\left(x^{2}-6x+9\right)\left(x-5\right)\times 2x$$

Apply the distributive property by multiplying each term of x^{2}-6x+9 by each term of x-5.

$$\left(x^{3}-5x^{2}-6x^{2}+30x+9x-45\right)\times 2x$$

Combine -5x^{2} and -6x^{2} to get -11x^{2}.

$$\left(x^{3}-11x^{2}+30x+9x-45\right)\times 2x$$

Combine 30x and 9x to get 39x.

$$\left(x^{3}-11x^{2}+39x-45\right)\times 2x$$

Use the distributive property to multiply x^{3}-11x^{2}+39x-45 by 2.

$$\left(2x^{3}-22x^{2}+78x-90\right)x$$

Use the distributive property to multiply 2x^{3}-22x^{2}+78x-90 by x.

$$2x^{4}-22x^{3}+78x^{2}-90x$$

Expand

$$2x^{4}-22x^{3}+78x^{2}-90x$$

Solution Steps

Multiply x-3 and x-3 to get \left(x-3\right)^{2}.

$$\left(x-3\right)^{2}\left(x-5\right)\times 2x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\left(x^{2}-6x+9\right)\left(x-5\right)\times 2x$$

Apply the distributive property by multiplying each term of x^{2}-6x+9 by each term of x-5.

$$\left(x^{3}-5x^{2}-6x^{2}+30x+9x-45\right)\times 2x$$

Combine -5x^{2} and -6x^{2} to get -11x^{2}.

$$\left(x^{3}-11x^{2}+30x+9x-45\right)\times 2x$$

Combine 30x and 9x to get 39x.

$$\left(x^{3}-11x^{2}+39x-45\right)\times 2x$$

Use the distributive property to multiply x^{3}-11x^{2}+39x-45 by 2.

$$\left(2x^{3}-22x^{2}+78x-90\right)x$$

Use the distributive property to multiply 2x^{3}-22x^{2}+78x-90 by x.

$$2x^{4}-22x^{3}+78x^{2}-90x$$