Answer
$$(4x^2-3x)(4x^2+3x)=16x^4-9x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4x^2-3x)(4x^2+3x)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}16x^4+12x^3-12x^3-9x^2 \xlongequal{ } \\[1 em] & \xlongequal{ }16x^4+ \cancel{12x^3} -\cancel{12x^3}-9x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}16x^4-9x^2\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4x^2-3x}\right) $ by each term in $ \left( 4x^2+3x\right) $. $$ \left( \color{blue}{4x^2-3x}\right) \cdot \left( 4x^2+3x\right) = 16x^4+ \cancel{12x^3} -\cancel{12x^3}-9x^2 $$ |
| ② | Combine like terms: $$ 16x^4+ \, \color{blue}{ \cancel{12x^3}} \, \, \color{blue}{ -\cancel{12x^3}} \,-9x^2 = 16x^4-9x^2 $$ |
This page was created using
Simplify polynomials calculator