 This video is called multiplying polynomials one there are four examples in this video Please watch and take notes on all four I know that this is a multiplication problem where this term has to multiply with this polynomial I know it's multiplication because there's nothing between the term and the parentheses There's no adding sign or subtracting sign, so I assume it's multiplication So what I'm gonna have to do is multiply this use my distributive property and multiply it by every term in the polynomial and please remember when multiplying terms like this you're gonna multiply the coefficients which are the numbers in front of your variable and you're gonna add the exponents so when I do negative 4y squared times a positive 5y to the fourth negative 4 times 5 is a negative 20 and then y squared times y to the fourth is y to the sixth. Next when I do a negative 4y squared times a negative 3y squared well the negative 4 times the negative 3 gives me a positive 12 and y squared times y squared is y to the fourth and then lastly when I do a negative 4y squared times 2 the negative 4 will multiply with the 2 to give me a minus 8 and then I just have y squared at the end. This answer is in standard form because I have an exponent of 6, 4 and 2 so from left to right it is going down. The second example I'm gonna be multiplying this negative 7h with everything inside my parentheses so I'm gonna have to do the distributive property 3 times and let's see what happens a negative 7h times a 3h squared is negative 21h to the third. You know that because the 7h when there's no exponent there there's really a 1 so 1 plus 2 becomes 3. Then the negative 7h minus or times a negative 8h will be a positive 56h squared because again there's really an exponent of 1 here and 1 plus 1 is 2 and then to finish I have to do a negative 7h times a negative 1. The negative 7 times negative 1 will be a positive 7 and then I'll have the h on the end. This is in standard form because from left to right the exponents are 3, 2 and 1 so from left to right they are going down. Alright example number 3. This one's a little bit different I've got two sets of parentheses and I know it's a multiplication problem because there's nothing in between the parentheses there's no adding sign or subtracting sign so I know it's multiplication. I'm gonna kind of have to do what I do in adding and subtracting polynomials is you are still combining like terms so this 1x to the 7th and 12x to the 4th will be multiplied together and the y to the 3rd and the y will be multiplied together. It's worth noting that the x to the 7th is really 1x to the 7th and here the y is really a 1y with an exponent of 1 as well so when I combine my like terms and multiply my x values I've got a 1x to the 7th and 12x to the 4th well 1 times 12 is 12 and then I add my exponents of 7 and 4 and get 11 so x to the 7th times 12x to the 4th is 12x to the 11th and then 1y to the 3rd times 1y to the 1st will be 1y to the 4th because 1 times 1 is 1 and then the exponent of 3 and exponent of 1 add to 4. Onto the last example I think I'm going to rewrite it quickly I've got 2 3rds x to the 4th k squared times 8x to the 5th k remember that k is really k to the 1st. I think what I'll do is remind myself that the x values are going to get multiplied together and then the k values are going to get multiplied together. It's worth noting that in front of the k is really a 1 in front of this k is really a 1 so when I multiply my x values together 2 3rds times 8 well I really get 16 over 3 so I have 16 over 3 x to the 9th because when I multiply my x's you add the exponents 4 plus 5 gave me that 9 then 1k squared times 1k to the 1st will be 1k to the 3rd sorry if you can hear my dog growling she wants to go out we have Sadie can you say hi okay so we've got 1k squared times 1k squared 1 times 1 is the 1 2 plus 1 is the 3