Special products and identities questions with answers

Questions

1. Question 1A bracket times a trinomial

   Expand and simplify $(x + 6)(2x^{2} + 6x + 7)$.

   Show the worked solution

   1. Step 1: Multiply each term of the first bracket by every term of the second$(x + 6)(2x^{2} + 6x + 7) = 2x^{3} + 6x^{2} + 7x + 12x^{2} + 36x + 42$
   2. Step 2: Collect like terms$= 2x^{3} + 18x^{2} + 43x + 42$
2. Question 2Products in two letters

   Expand and simplify $2x^{2} - (x + 2y)(x - 2y)$.

   Show the worked solution

   1. Step 1: The product is a difference of two squares: (A + B)(A − B) = A² − B²$2x^{2} - (x + 2y)(x - 2y) = 2x^{2} - (x^{2} - 4y^{2})$
   2. Step 2: Remove the bracket (the minus changes every sign inside) and collect like terms$= x^{2} + 4y^{2}$
3. Question 3Values from the identities

   Given that $(m + n)^{2} = 4$ and $mn = -48$, find the value of $m^{2} + n^{2}$.

   Show the worked solution

   1. Step 1: Write m² + n² using what you are given$m^{2} + n^{2} = (m + n)^{2} - 2mn$
   2. Step 2: Substitute the values given$= (4)^{2} - 2(-48)$
   3. Step 3: Work it out$= 100$
4. Question 4a² − b² with numbers

   Without using a calculator, evaluate $44^{2} - 6^{2}$.

   Show the worked solution

   1. Step 1: Use a² − b² = (a + b)(a − b)$44^{2} - 6^{2} = (44 + 6)(44 - 6)$
   2. Step 2: Work out the brackets, then multiply$= 50 \times 38$$= 1900$
5. Question 5Show it is a multiple of k

   Show that $(4n + 4)(4n - 4) - (4n - 4)^{2}$ is a multiple of $32$ for every positive integer $n$.

   Show the worked solution

   1. Step 1: Expand the product and the square$(4n + 4)(4n - 4) - (4n - 4)^{2} = (16n^{2} - 16) - (16n^{2} - 32n + 16)$
   2. Step 2: Remove the brackets (the minus changes every sign in the second) and collect like terms$= 32n - 32$
   3. Step 3: Take out the common factor 32$= 32(n - 1)$$n - 1$ is a whole number, so this is a multiple of $32$ for every positive integer $n$.
6. Question 6Why it is always even, odd or a multiple

   $n$ is a positive integer. Which statement explains why $n^{2} + 3n$ is always even?

   Show the worked solution

   1. Step 1: Factorise$n^{2} + 3n = n(n + 3)$
   2. Step 2: What does that form tell you?It is $n(n + 3)$: if $n$ is odd then $n + 3$ is even, so one factor is even.