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yulyashka [42]
2 years ago
14

Using the distributive property to find the product (y−4x)(y2+4y+16) results in a polynomial of the form y3+4y2+ay−4xy2−axy−64x.

What is the value of a in the polynomial?
Mathematics
1 answer:
Alla [95]2 years ago
8 0
<h3>Answer:  a = 16</h3>

==========================================================

Explanation:

Let's expand out that given expression.

First I'll let w = y-4x to make distribution a bit easier

(y-4x)(y^2+4y+16)\\\\w(y^2+4y+16)\\\\wy^2+4wy+16w\\\\

Now plug w = y-4x back in and distribute three more times

wy^2+4wy+16w\\\\y^2(w)+4y(w)+16(w)\\\\y^2(y-4x)+4y(y-4x)+16(y-4x)\\\\y^3-4xy^2+4y^2-\boldsymbol{16}xy+16y-64x\\\\

Notice that the only xy term here is the -16xy

Comparing this to the form -axy shows that \boldsymbol{a = 16}

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Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
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--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
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