I’ve said it before and I’ll say it again: the writers of standardized tests assess the same concepts, over and over and over again. Test experts are able to become test experts for that very reason—there is a finite amount of material one needs to learn in order to master the ACT and the SAT. If you study a few dozen tests and find these predictable patterns, you could be an expert, too.

Luckily, I’m here to save you from that mind-numbing task. And trust me—it’s mind-numbing, especially when we get into some of the concepts from Algebra II, such as classic quadratic forms.

1. (*x* + *y*)(*x* – *y*) = *x*^{2} – *y*^{2}

Examples:

(*t* + 5)(*t* – 5) = *t*^{2} – 5^{2} = *t*^{2} – 25

(3*a* + *b*)(3*a* – *b*) = (3*a*)^{2} – *b*^{2} = 9*a*^{2} – *b*^{2}

*y*^{2} – 64 = (*y* + 8)(*y* – 8)

36 – *n*^{2 }= (6 + *n*)(6 – *n*)

2. (*x* + *y*)^{2} = *x*^{2} + 2*xy* + *y*^{2}

Examples:

(*t* + 5)^{2} = *t*^{2} + 2(*t*)(5) + 5^{2} = *t*^{2} + 10*t* + 25

(3*a* + *b*)^{2} = (3*a*)^{2} + 2(3*a*)(*b*) + *b*^{2} = 9*a*^{2} + 6*ab* + *b*^{2}

(6 + *n*)^{2}* = *6^{2} + 2(6)(*n*) + *n*^{2 }*= *36 + 12*n* + *n*^{2}

3. (*x* – *y*)^{2} = *x*^{2} – 2*xy* + *y*^{2}

Examples:

(*t* – 5)^{2} = *t*^{2} – 2(*t*)(5) + 5^{2} = *t*^{2} – 10*t* + 25

(3*a* – *b*)^{2} = (3*a*)^{2} – 2(3*a*)(*b*) + *b*^{2} = 9*a*^{2} – 6*ab* + *b*^{2}

(6 – *n*)^{2}* = *6^{2} – 2(6)(*n*) + *n*^{2 }*= *36 – 12*n* + *n*^{2}

Now let’s look at how memorizing these forms can help you quickly and confidently solve an ACT or SAT question.

1. If

r–p= 7, what is the value ofr^{2}– 2rp+p^{2}?(A) 7

(B) 14

(C) 28

(D) 35

(E) 49

A student who doesn’t know the classic ACT and SAT quadratic equations will start to (unsuccessfully) use substitution:

r – p = 7

r = 7 + p

They will plug *7 + p* into the second expression for *r*, even though the second expression is just an expression—not an equation!

r^{2} – 2rp + p^{2}

(7 + p)^{2} – 2(7 + p)p + p^{2}

(7 + p) (7 + p) – 2(7 + p)p + p^{2}

49 + 14p + 2p^{2} – 2(7p + p^{2}) + p^{2}

49 + 14p + 2p^{2} – 14p – 2p^{2} + p^{2}

49 + 2p^{2}– 2p^{2} + p^{2}

49 + p^{2}

Oh, wait! It’s just an expression. It doesn’t equal anything. This won’t work!

A lot of wasted time for the average student.

But for you, the reader of great test prep blogs, this question is solved in about 5 seconds. You look at *r ^{2} – 2rp + p^{2}* and immediately recognize the third classic quadratic form:

(*x* – *y*)^{2} = *x*^{2} – 2*xy* + *y*^{2}

So *(r – p) ^{2} *=

*r*

^{2}– 2rp + p^{2}And since *r – p = 7*, then *(r – p) ^{2} *=

*7*

^{2}= 49.Voila.

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