How do you find the degree of $y = (\frac{1}{4}) {(t - 1)}^{2} (t + 3) (4 - t)$ $y = (1/4)(t-1)^2(t+3)(4-t)$ ?

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Answer 1 · 2018-06-07T05:38:07

$y$ $y$ is of degree $4$ $4$

The degree of a polynomial of a single variable is the value of the highest exponent of the variable.

In our example:

$y = (\frac{1}{4}) {(t - 1)}^{2} (t + 3) (4 - t)$ $y = (1/4)(t-1)^2(t+3)(4-t)$

In this case the variable is $t$ $t$

We could go to the bother of expanding $y$ $y$ to find the highest exponent of $t$ $t$ . However, in this case there is a much simpler way.

Since $y$ $y$ is the product of terms we can simply find the degree of each term and sum each to find the degree of $y$ $y$ .

Taking each term in turn:

$\frac{1}{4} = \frac{1}{4} t^{0} \to$ $1/4 = 1/4t^0 ->$ Degree 0

${(t - 1)}^{2} \to$ $(t-1)^2 ->$ Degree 2

$(t + 3) \to$ $(t+3) ->$ Degree 1

$(4 - t) \to$ $(4-t) ->$ Degree 1

Hence, degree of $y = 0 + 2 + 1 + 1 = 4$ $y = 0+2+1+1 =4$

NB: This only works because $y$ $y$ is the product of terms.

We are actually using the property of exponents:

$t^{a} \times t^{b} = t^{a + b}$ $t^a xx t^b = t^(a+b)$

How do you find the degree of y = (1/4)(t-1)^2(t+3)(4-t)y=(14)(t−1)2(t+3)(4−t)y = (1/4)(t-1)^2(t+3)(4-t)?