f(x)=(16x^4-81)(2x-3) −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 −3 -2.5 −2 -1.5 −1 -0.5 0 0.5 1 1.5 2 2.5 3 x y
Exercise( 16 x 4 − 81 ) ( 2 x − 3 ) \left(16x^4-81\right)\left(2x-3\right) ( 16 x 4 − 81 ) ( 2 x − 3 )
Step-by-step Solution
1
Multiply the single term 2 x − 3 2x-3 2 x − 3 ( 16 x 4 − 81 ) \left(16x^4-81\right) ( 16 x 4 − 81 )
16 x 4 ( 2 x − 3 ) − 81 ( 2 x − 3 ) 16x^4\left(2x-3\right)-81\left(2x-3\right) 16 x 4 ( 2 x − 3 ) − 81 ( 2 x − 3 )
2
Multiply the single term 16 x 4 16x^4 16 x 4 ( 2 x − 3 ) \left(2x-3\right) ( 2 x − 3 )
32 x ⋅ x 4 − 48 x 4 − 81 ( 2 x − 3 ) 32x\cdot x^4-48x^4-81\left(2x-3\right) 32 x ⋅ x 4 − 48 x 4 − 81 ( 2 x − 3 )
Intermediate steps
3
When multiplying exponents with same base you can add the exponents: 32 x ⋅ x 4 32x\cdot x^4 32 x ⋅ x 4
32 x 5 − 48 x 4 − 81 ( 2 x − 3 ) 32x^{5}-48x^4-81\left(2x-3\right) 32 x 5 − 48 x 4 − 81 ( 2 x − 3 )
Explain this step further
4
Multiply the single term − 81 -81 − 81 ( 2 x − 3 ) \left(2x-3\right) ( 2 x − 3 )
32 x 5 − 48 x 4 − 162 x + 243 32x^{5}-48x^4-162x+243 32 x 5 − 48 x 4 − 162 x + 243
Final answer to the exercise 32 x 5 − 48 x 4 − 162 x + 243 32x^{5}-48x^4-162x+243 32 x 5 − 48 x 4 − 162 x + 243