Expand and simplify seven minus
three minus 𝑦, 𝑦 plus two.
In this question, we can see we
have two binomials which are multiplied together. We’re going to expand these
binomials first and then subtract the answer from seven. We’re going to use the grid method
to multiply these binomials. And notice the negative in front of
them isn’t included.
Setting up our grid, we have three
minus 𝑦 as our first binomial. Our second binomial can be split
into the terms 𝑦 and two. And we can write that with or
without the plus sign. It doesn’t matter which way round
do we put our binomials. Filling in our grid, we start with
𝑦 times three, which is three 𝑦. Next, we have 𝑦 times negative 𝑦,
which will give us negative 𝑦 squared. On the next row, we have two times
three, which is six. And the final term is calculated by
two times negative 𝑦, which is negative two 𝑦.
To take our answer from the grid
then, we add together the four products we’ve just calculated. Starting with our largest exponent
of 𝑦, we have negative 𝑦 squared. Next, we notice that we have two
terms in 𝑦, which we can collect together. So three 𝑦 plus negative two 𝑦
will give us plus 𝑦. And then, we add on our final term
from the grid which is plus six. So to answer the question then of
seven minus three minus 𝑦, 𝑦 plus two, we replace what we’ve worked out in our
expansion, giving us seven minus minus 𝑦 squared plus 𝑦 plus six.
It’s important to include all of
these in parentheses since we want to subtract all of these terms. It can be helpful to write the next
line where we distribute the negative across all the terms. So we have seven minus minus 𝑦
squared, which is plus 𝑦 squared. A negative times plus 𝑦 will give
us negative 𝑦. And finally, a negative times plus
six will give us negative six.
To simplify then, we check if there
are any like terms. We can see that we have a seven and
a negative six, which is equivalent to one, giving us a final answer of 𝑦 squared
minus 𝑦 plus one. We could write these terms in any
order. But by convention, we usually write
them in decreasing exponent values. Here, we have our 𝑦 squared first,
then our 𝑦 term, and then our constant.