Solve f(x)=(3x+4)^2(x-7)(2x-5)^3

$$f ( x ) = ( 3 x + 4 ) ^ { 2 } ( x - 7 ) ( 2 x - 5 ) ^ { 3 }$$

Evaluate

$$\left(x-7\right)\left(3x+4\right)^{2}\left(2x-5\right)^{3}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$\left(9x^{2}+24x+16\right)\left(x-7\right)\left(2x-5\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(2x-5\right)^{3}.

$$\left(9x^{2}+24x+16\right)\left(x-7\right)\left(8x^{3}-60x^{2}+150x-125\right)$$

Use the distributive property to multiply 9x^{2}+24x+16 by x-7 and combine like terms.

$$\left(9x^{3}-39x^{2}-152x-112\right)\left(8x^{3}-60x^{2}+150x-125\right)$$

Use the distributive property to multiply 9x^{3}-39x^{2}-152x-112 by 8x^{3}-60x^{2}+150x-125 and combine like terms.

$$72x^{6}-852x^{5}+2474x^{4}+1249x^{3}-11205x^{2}+2200x+14000$$

Expand

$$72x^{6}-852x^{5}+2474x^{4}+1249x^{3}-11205x^{2}+2200x+14000$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$\left(9x^{2}+24x+16\right)\left(x-7\right)\left(2x-5\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(2x-5\right)^{3}.

$$\left(9x^{2}+24x+16\right)\left(x-7\right)\left(8x^{3}-60x^{2}+150x-125\right)$$

Use the distributive property to multiply 9x^{2}+24x+16 by x-7 and combine like terms.

$$\left(9x^{3}-39x^{2}-152x-112\right)\left(8x^{3}-60x^{2}+150x-125\right)$$

Use the distributive property to multiply 9x^{3}-39x^{2}-152x-112 by 8x^{3}-60x^{2}+150x-125 and combine like terms.

$$72x^{6}-852x^{5}+2474x^{4}+1249x^{3}-11205x^{2}+2200x+14000$$