Multiply the polynomials.(x+3)(x−5)(−2x+1)\left(x+3\right)\left(x-5\right)\left(-2x+1\right)

− 2 x 3 + 5 x 2 + 28 x − 15 -2x^3+5x^2+28x-15 − 2 x 3 + 5 x 2 + 28 x − 15

Multiply the polynomials. ( x + 3 ) ( x − 5 ) ( − 2 x + 1 ) \left(x+3\right)\left(x-5\right)\left(-2x+1\right)

Hey everyone. So let's take a look at another practice problem here. So here I've got 3 polynomials that are multiplied together. X plus 3, x minus 5, and the negative 2 x plus 1. How do I multiply all these three things together? Well, basically, what happens is you can break it up into a couple of steps. We're gonna multiply these 2 polynomials because these 2 are relatively simple binomials. So when we do this expression over here and we multiply this because it's a 2 times a 2 term expression, we can use the foil method. Then later on, we can sort of bring down this negative 2 x plus 1, and we'll deal with that later. So let's just do the foil method from here for for these 2, first. So remember the foil method is we're gonna do first times the first, so this just becomes x squared. Then we'll do the outer terms, x times negative five becomes negative 5x. The inner terms becomes 3 and x, this is plus 3x. And the last terms is the 3 and the negative 5, and that's minus 15. Now we just do the simplification. This just becomes I get x squared minus 5x. Oops. Actually, the negative 5x and the 3x combined to negative 2x. Right? Negative 5 plus 3, and then I've got negative 15 over here. So this is what I've gotten out of the foil. Am I done yet? No. Because you still haven't multiplied this negative two x squared or negative two x plus one term over here. So because we haven't done that yet, now what we basically have to do is we have to bring this down over here, and this just becomes negative two x plus 1. And we're gonna multiply that by the expression that we just got by foiling. So you have to break it up into 2 steps. 1st, you're gonna foil these 2, then you're gonna take this expression and multiply it by this one. So the question is, can I just foil these 2? And remember, you can't do that because now we've ended up with because we ended up with a 3 term expression multiplied by a 2 term expression. So we're gonna have to do this thing where we break them off into 2 different pieces. So here it tells how this is gonna work. Take the shortest expression and break that up. So really what happens is this becomes negative 2x on the outside, and then I'm gonna put something in parentheses plus the one on the outside, and and that's just gonna go in the parentheses over here. So here what I've got is I just basically put this expression inside of both these parentheses. So this is x squared minus 2x minus 15, and then over here I got x squared minus 2x minus 15. So now what I'm gonna go ahead and do is now I have to do an extra step, and I have to distribute this term into all of these terms over here. I'll do that in blue. So negative 2, remember there's a negative out here, which is really important, and you have to distribute that. This is negative two x times x squared. So this first term here just becomes negative 2x cubed. Right? So negative 2x into negative 2x, I'm gonna multiply negative 2x times negative 2x. The 2 and the 2 become 4, the negatives cancel, and the x and the x's become x squared. So this becomes 4x squared. And then -2x and the -fifteen, again the negatives will cancel, so you'll get 30x. That's what you get from the first expression. What do you get from the second one? Well, this one's actually relatively straightforward because all you're doing is you're just multiplying 1 by every one of these terms, so it actually doesn't really change anything here. All you're really doing here is you could just add the x squared minus the 2x minus the 15. The one distributed into all of these terms doesn't actually do anything. Okay? So now what we can do here is now we can do the last step, which is going which is gonna be us combining all the like terms. Alright? So let's do this look through this expression. We have a couple of x cubes or we I see some x cubes, some x squared. So this is my x cube terms over here. I have nothing else. I've got some x squared terms over here. I've got the 4x and the plus x over here. I've also got some, some, you know, some x's over here. So I've got plus 30x and negative 2x. And then finally, I've also got some some constants over here as well. Alright? So let's just go ahead and do this. The negative two x squared doesn't change. So this is negative two x squared. Then I have, plus four x squared and then 1x2. So this combines down to 5x2, these two terms over here. Then I've got the 30x and negative 2x that combines to 28x, that's what those two terms become, and then the -fifteen doesn't combine with anything. So this expression over here, let me just scoot this down, is actually our most simplified because I can't combine any more like terms. So this is our final answer. Alright? So in these types of problems where you have 3 things multiplied, you have to foil the first term, and then you're gonna have to multiply using this distributive property with that second expression. Alright? So it's a little bit tedious, but, you know, you'll see that it's pretty straightforward. Hopefully, I got the right answer. Thanks for

Multiply the polynomials. ( x + 3 ) ( x − 5 ) ( − 2 x + 1 ) \left(x+3\right)\left(x-5\right)\left(-2x+1\right)

− 2 x 3 + 5 x 2 + 28 x − 15 -2x^3+5x^2+28x-15 − 2 x 3 + 5 x 2 + 28 x − 15

− 2 x 3 + 4 x 2 + 30 x -2x^3+4x^2+30x

2 x 3 + 5 x 2 + 28 x − 15 2x^3+5x^2+28x-15 2 x 3 + 5 x 2 + 28 x − 15

6. Exponents, Polynomials, and Polynomial Functions

Multiplying Polynomials

Intermediate Algebra

6. Exponents, Polynomials, and Polynomial Functions

Multiplying Polynomials

Struggling with Intermediate Algebra?

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Multiple Choice

Multiply the polynomials.
$(x + 3) (x - 5) (- 2 x + 1)$

$x^{2} - 2 x - 15$

$- 2 x^{3} + 4 x^{2} + 30 x$

$-2x^3+5x^2+28x-15$

$2x^3+5x^2+28x-15$

Verified step by step guidance

First, multiply the first two binomials $\left(x+3\right)$ and $\left(x-5\right)$ using the distributive property (FOIL method): multiply each term in the first binomial by each term in the second binomial.

Write out the products explicitly: $x \cdot x$, $x \cdot (-5)$, \$3 \cdot x$, and $3 \cdot (-5)$, then combine like terms to simplify the resulting quadratic polynomial.

Next, take the simplified quadratic polynomial from step 2 and multiply it by the third binomial $\left(-2x+1\right)$, again using the distributive property: multiply each term of the quadratic by each term of the binomial.

Expand all products carefully, making sure to multiply coefficients and add exponents for like bases (e.g., $x^m \cdot x^n = x^{m+n}$), and write down all resulting terms.

Finally, combine all like terms (terms with the same power of $x$) to write the polynomial in standard form, ordered from the highest degree term to the constant term.

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