\(\displaystyle Expand \; the \; following: \;x(x-2)\)\(\displaystyle Distribute \; x \; over \; x-2.\)\(\displaystyle xx+x(-2)\)
\(\displaystyle Answer: \; x^2-2x\)

$\displaystyle ab(ab^{2}+4a-b)$

Using the distributive property you are simply going to share the term $\displaystyle ab$, with every term in the poynomial

$\displaystyle ab\cdot ab^{2}+ ab\cdot 4a- ab\cdot b$

Now because we are multiplying like variables we can add the exponents, to simplify each expression

$\displaystyle a^{2}b^{3}+4a^{2}b-ab^{2}$   

This will be our final answer because we can not add terms unless they are 'like' meaning they contain the same elements and degrees. 

$\displaystyle -3t ^{2} ( 2t ^{2} - 5t + 7)$

$\displaystyle = -3t ^{2} \cdot 2t ^{2} - \left (-3t ^{2} \right )\cdot 5t +\left (-3t ^{2} \right )\cdot 7$

$\displaystyle = -3t ^{2} \cdot 2t ^{2} + 3t ^{2} \cdot 5t -3t ^{2}\cdot 7$

$\displaystyle = -3\cdot 2 \cdot t ^{2} \cdot t ^{2} + 3\cdot 5 \cdot t ^{2} \cdot t -3\cdot 7 \cdot t ^{2}$

$\displaystyle = -6 \cdot t ^{2+2} + 15\cdot t ^{2+1} -21 \cdot t ^{2}$

$\displaystyle = -6 t ^{4} + 15t ^{3} -21t ^{2}$

Distribute the $\displaystyle 2p^{2}$ to each term in the parentheses in the other polynomial.

$\displaystyle 2p^{2}(4x)=8xp^{2}$

$\displaystyle 2p^{2}(p)=2p^{3}$

Putting the results back together

$\displaystyle (2p^{2})(4x+p)=8xp^{2}+2p^{3}$

Multiply each term of the polynomial by $\displaystyle -4x$. Be careful to distribute the negative sign.

$\displaystyle (-4x)(3x^2)=-12x^3$

$\displaystyle (-4x)(-2x)=8x^2$

$\displaystyle (-4x)(4)=-16x$

Add the individual terms together:

$\displaystyle -12x^3+8x^2+(-16x)=12x^3+8x^2-16x$

Distribute $\displaystyle 7y$ to each term in the parentheses in the polynomial

$\displaystyle 7y(7t)=49yt$

$\displaystyle 7y(-3y^{3})=-21y^{4}$

$\displaystyle 7y(2)=14y$

Combine the results

$\displaystyle (7y)(7t-3y^{3} + 2)=49yt-21y^{4} +14y$

$\displaystyle \small (x-4)(x+2)(2x-5)$

When multiplying, the order in which you multiply does not matter. Let's start with the first two monomials.

 $\displaystyle (x-4)(x+2)$

Use FOIL to expand.

$\displaystyle x^2+2x-4x-8=x^2-2x-8$

Now we need to multiply the third monomial.

$\displaystyle \small (x-4)(x+2)(2x-5)=(x^2-2x-8)(2x-5)$

Similar to FOIL, we need to multiply each combination of terms.

$\displaystyle 2x(x^2-2x-8)+(-5)(x^2-2x-8)$

$\displaystyle 2x^3-4x^2-16x-5x^2+10x+40$

Combine like terms.

$\displaystyle 2x^3-9x^2-6x+40$

First, mulitply the mononomial by the first term of the polynomial:

$\displaystyle \small 7n\times8n\ = 56n^2$

Second, multiply the monomial by the second term of the polynomial:

$\displaystyle \small 7n\times (-2)\ = -14n$

Add the terms together:

$\displaystyle \small 56n^2\ +\ (-14n)\ = 56n^2-14n$

To expand, multiply 8x by both terms in the expression (3x + 7).

8x multiplied by 3x is 24x².

8x multiplied by 7 is 56x.

Therefore, 8x(3x + 7) = 24x² + 56x.

We need to distribute the 4x² through the terms in the parentheses:

$\displaystyle 4x^2(3x^2+2x+4)=4x^2(3x^2)+4x^2(2x)+4x^2(4)$

This becomes $\displaystyle 12x^4+8x^3+16x^2$.