Question

use pascals triangle to complete the expansion of $(y - z)^4$.
$y^4 + \square y^3 z + 6y^2 z^2 + \square y z^3 + z^4$

Explanation:

Step1: Recall Pascal's Triangle for \(n = 4\)

Pascal's Triangle for the 4th power (since the exponent is 4) has coefficients \(1, 4, 6, 4, 1\). The general form of \((a + b)^n\) using Pascal's Triangle is \(\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). For \((y - z)^4=(y+(-z))^4\), the coefficients from Pascal's Triangle (row \(n = 4\)) are \(1, 4, 6, 4, 1\).

Step2: Determine the coefficient for \(y^{3}z\)

In the expansion of \((y - z)^4\), the term with \(y^{3}z\) corresponds to \(k = 1\) (since \(y^{4 - 1}(-z)^{1}=y^{3}(-z)\)). The coefficient from Pascal's Triangle for \(k = 1\) (and considering the sign: \((-z)^1=-z\), but let's look at the given form. The given form has \(y^{3}z\) (not \(-y^{3}z\) yet? Wait, no, let's check again. Wait, the expansion of \((y - z)^4\) is \(y^{4}-4y^{3}z + 6y^{2}z^{2}-4yz^{3}+z^{4}\). But the given form has \(y^{4}+\square y^{3}z + 6y^{2}z^{2}+\square yz^{3}+z^{4}\). Wait, maybe there's a sign error in the problem's given form? Wait, no, maybe the problem is written as \((y + (-z))^4\), but when expanding, the coefficients from Pascal's Triangle are \(1, 4, 6, 4, 1\), but with signs alternating. Wait, no, let's re - express \((y - z)^4=(y+(-z))^4\). Using the binomial theorem, \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(a = y\), \(b=-z\). So for \(k = 1\): \(\binom{4}{1}y^{3}(-z)^{1}=4y^{3}(-z)=- 4y^{3}z\). But the given form has \(+\square y^{3}z\), so maybe there's a mistake in the problem's sign, or maybe we are to consider the absolute value of the coefficient? Wait, no, let's check the Pascal's Triangle row for \(n = 4\): \(1, 4, 6, 4, 1\). The expansion of \((y - z)^4\) is \(y^{4}-4y^{3}z + 6y^{2}z^{2}-4yz^{3}+z^{4}\). But the given form is \(y^{4}+\square y^{3}z + 6y^{2}z^{2}+\square yz^{3}+z^{4}\). So to match the form, the coefficient for \(y^{3}z\) should be \(- 4\)? But the box is for a number, maybe the problem has a typo, but if we look at the Pascal's Triangle coefficients (ignoring the sign for the moment, but that's wrong). Wait, no, maybe the problem is actually \((y + z)^4\) but written as \((y - z)^4\) by mistake? No, the problem says \((y - z)^4\). Wait, let's recast:

Wait, the expansion of \((y - z)^4\) is:

\[ \begin{align*} (y - z)^4&=\binom{4}{0}y^{4}(-z)^{0}+\binom{4}{1}y^{3}(-z)^{1}+\binom{4}{2}y^{2}(-z)^{2}+\binom{4}{3}y^{1}(-z)^{3}+\binom{4}{4}y^{0}(-z)^{4}\\ &=y^{4}-4y^{3}z + 6y^{2}z^{2}-4yz^{3}+z^{4} \end{align*} \]

But the given form is \(y^{4}+\square y^{3}z + 6y^{2}z^{2}+\square yz^{3}+z^{4}\). So if we compare, the coefficient of \(y^{3}z\) in the correct expansion is \(- 4\), but the given form has a plus sign. Maybe the problem has a sign error, and we are to put the absolute value of the coefficient? Wait, no, let's check the next term. The term with \(yz^{3}\) in the correct expansion is \(-4yz^{3}\), and the given form has \(+\square yz^{3}\). So maybe the problem is written as \((y + z)^4\) by mistake. If we consider \((y + z)^4\), the expansion is \(y^{4}+4y^{3}z + 6y^{2}z^{2}+4yz^{3}+z^{4}\). So in that case, the coefficient for \(y^{3}z\) is \(4\) (but with a sign error in the problem's \((y - z)^4\) to \((y + z)^4\) - like form). Similarly, the coefficient for \(yz^{3}\) is \(4\) (but with a sign error). So, assuming that the problem has a typo and is actually expanding \((y + z)^4\) (or is ignoring the sign of \(z\) in the terms), the coefficient for \(y^{3}z\) is \(4\) (but with a negative sign in the correct \((y - z)^4\) expansion) and the coefficient for \(yz^{3}\) is \(4\) (with a negative sign in the correct expansion). But looking at the given form, the first blank is for the coefficient of \(y^{3}z\) and the second for \(yz^{3}\). From Pascal's Triangle, the row for \(n = 4\) is \(1, 4, 6, 4, 1\). So for the expansion of \((a + b)^4=a^{4}+4a^{3}b + 6a^{2}b^{2}+4ab^{3}+b^{4}\). If we let \(a = y\) and \(b=-z\), then \((y - z)^4=y^{4}+4y^{3}(-z)+6y^{2}(-z)^{2}+4y(-z)^{3}+(-z)^{4}=y^{4}-4y^{3}z + 6y^{2}z^{2}-4yz^{3}+z^{4}\). But the given form has \(+\square y^{3}z\) and \(+\square yz^{3}\). So there is a sign mismatch. However, if we consider the problem's given form (maybe a mistake in the sign of \(z\) in the problem statement), and just take the coefficients from Pascal's Triangle (ignoring the sign of \(z\) for the coefficient calculation), then the coefficient for \(y^{3}z\) is \(4\) (but with a negative sign in reality) and for \(yz^{3}\) is \(4\) (but with a negative sign in reality). But since the problem's form has \(+\square y^{3}z\) and \(+\square yz^{3}\), maybe the problem intended to expand \((y + z)^4\) instead of \((y - z)^4\). In that case, the coefficients are \(4\) and \(4\).

Step3: Determine the coefficient for \(yz^{3}\)

For the term \(yz^{3}\) in the expansion of \((y - z)^4\), it corresponds to \(k = 3\) (since \(y^{4 - 3}(-z)^{3}=y(-z)^{3}\)). The binomial coefficient \(\binom{4}{3}=4\), and \((-z)^{3}=-z^{3}\), so the term is \(4y(-z^{3})=-4yz^{3}\). But in the given form, it's \(+\square yz^{3}\), so again, if we take the coefficient from Pascal's Triangle (ignoring the sign of \(z\)), it's \(4\).

Answer:

The coefficient for \(y^{3}z\) is \(- 4\) (but if we follow the problem's form which has a plus sign, maybe there's a mistake, but if we consider the Pascal's Triangle coefficients for the absolute values, the first blank is \(-4\) and the second blank is \(-4\). Wait, no, let's re - evaluate. The expansion of \((y - z)^4\) is:

$$ (y - z)^4=y^{4}-4y^{3}z + 6y^{2}z^{2}-4yz^{3}+z^{4} $$

Comparing with the given form \(y^{4}+\square y^{3}z + 6y^{2}z^{2}+\square yz^{3}+z^{4}\), we can see that the coefficient of \(y^{3}z\) should be \(-4\) (so that \(+\square y^{3}z=-4y^{3}z\)) and the coefficient of \(yz^{3}\) should be \(-4\) (so that \(+\square yz^{3}=-4yz^{3}\)).

So the first blank (coefficient of \(y^{3}z\)) is \(-4\) and the second blank (coefficient of \(yz^{3}\)) is \(-4\).