Question

study the products shown. is there a pattern?
$(x + 3)^2 = x^2 + 6x + 9$
$(x + 4)^2 = x^2 + 8x + 16$
$(x + 5)^2 = x^2 + 10x + 25$
$(x + 6)^2 = x^2 + 12x + 36$

Explanation:

Step1: Analyze the general form of \((x + a)^2\)

The formula for squaring a binomial is \((x + a)^2=x^{2}+2ax + a^{2}\).

Step2: Compare with given examples

• For \((x + 3)^2\), here \(a = 3\). Then \(2ax=2\times3\times x = 6x\) and \(a^{2}=3^{2}=9\), so \((x + 3)^2=x^{2}+6x + 9\).
• For \((x + 4)^2\), here \(a = 4\). Then \(2ax=2\times4\times x=8x\) and \(a^{2}=4^{2} = 16\), so \((x + 4)^2=x^{2}+8x + 16\).
• For \((x + 5)^2\), here \(a=5\). Then \(2ax = 2\times5\times x=10x\) and \(a^{2}=5^{2}=25\), so \((x + 5)^2=x^{2}+10x + 25\).
• For \((x + 6)^2\), here \(a = 6\). Then \(2ax=2\times6\times x = 12x\) and \(a^{2}=6^{2}=36\), so \((x + 6)^2=x^{2}+12x + 36\).

We can see that when we square the binomial \((x + a)\) (where \(a\) is \(3,4,5,6\) in the given examples), the result follows the pattern \((x + a)^2=x^{2}+2ax + a^{2}\), where the middle term is \(2\) times the product of \(x\) and the constant term in the binomial, and the last term is the square of the constant term in the binomial.

Answer:

The pattern is that for a binomial \((x + a)^2\) (where \(a\) is a constant), the expansion is \(x^{2}+2ax + a^{2}\). In the given examples, when \(a = 3,4,5,6\), we have \((x + 3)^2=x^{2}+6x + 9\) (since \(2\times3 = 6\) and \(3^{2}=9\)), \((x + 4)^2=x^{2}+8x + 16\) (since \(2\times4=8\) and \(4^{2} = 16\)), \((x + 5)^2=x^{2}+10x + 25\) (since \(2\times5 = 10\) and \(5^{2}=25\)), \((x + 6)^2=x^{2}+12x + 36\) (since \(2\times6=12\) and \(6^{2}=36\)). So the pattern follows the perfect - square binomial formula \((x + a)^2=x^{2}+2ax + a^{2}\).