A perfect square trinomial is a trinomial that can be written as the square of a binomial. There are two general types of perfect square trinomials that can be useful when dealing with quadratic functions or quadratic equations.
Perfect Square Trinomial | Square of Binomial |
---|---|
a2+2ab+b2 | (a+b)2 |
a2−2ab+b2 | (a−b)2 |
The expressions in the table are perfect square trinomials. Complete the expressions to ensure equivalence between corresponding standard forms and factored forms.
standard form | factored form |
---|---|
x2−2x+1 | (x__ ___)2 |
x2−___x+___ | (x−7)2 |
x2−10x+___ | (x__ ___)2 |
x2+___x+100 | (x__ ___)2 |
Here the same expressions are written both in standard form and in factored form, but one or more terms are missing in each expression. To identify the missing term(s) the rules for factoring a perfect square trinomial are useful.
To help us identify the terms in the rule in the first example, we can for the expression x2−2x+1 rewrite 1 as 12 and 2x as 2⋅x⋅1.
To make the expressions match each other we will fill in −1 where the blanks are.
We can use this reasoning to complete the other perfect square trinomials in the table. Next, we'll consider the second row.
To fill in the second blank we use that b must be 7, which gives us that b2=72. When we study the first blank we can match a with x. What is left is 2⋅b=2⋅7.
This we prefer writing as x2−14x+49=(x−7)2. Let's continue with the third row
Here the terms a and x match each other. Since the terms 2ab and 10x must match each other we find that 2b=10, or b=5. Let's use this to fill in the blanks in the expression.
By squaring 5 we get that the second row reads x2−10x+25=(x−5)2. We are now going to deal with the last row in the table.
The lines match each other when b=10, giving us the first blank 2⋅10=20 and the second blank +10. Let's write them together with the blanks filled in.
We'll summarize our results by adding the found values to the table.
standard form | factored form |
---|---|
x2−2x+1 | (x−1)2 |
x2−14x+49 | (x−7)2 |
x2−10x+25 | (x−5)2 |
x2+20x+100 | (x+10)2 |
Split into factors
Commutative Property of Multiplication
a2+2ab+b2=(a+b)2
Complete the square
a2−2ab+b2=(a−b)2
Subtract term
LHS+1=RHS+1
Rearrange equation
LHS=RHS
LHS+0.5=RHS+0.5
State solutions