Question

Which polynomial is a perfect square trinomial? (4 points)

A 25x2 - 20x - 16

B 9a2 - 30a - 25

C 25b2 - 15b + 9

D 16x2 - 24x + 9

Answer 1

which options can you exclude right away ?

Answer 2

erm... D?

Answer 3

The last term, when factoring ANY expression \(\large\color{slate}{ (x-a)^2 }\) should [always] be negative or positive ?

Answer 4

\(\large\color{slate}{ (x-a)^2~~\Rightarrow~~x^2-2ax+a^2 }\)

Answer 5

positive

Answer 6

yes

Answer 7

(and certainly \(\large\color{slate}{ (x+a)^2 }\) wouldn't have any negative -- which implies for sure no negative last term)

Answer 8

So which options do you have left /

Answer 9

so A and B are thrown out.

Answer 10

c and d

Answer 11

yes, very good!

Answer 12

C. \(\large\color{slate}{ 25b^2 - 15b + 9 }\)

D. \(\large\color{slate}{ 16x^2 - 24x + 9 }\)

Answer 13

C?

Answer 14

always keep in mind:
\(\large\color{blue}{ (A-B) ^2=A^2-2AB+B^2}\)

For C,

\(\large\color{slate}{ 25b^2 - 15b + 9 }\)
\(\large\color{slate}{ (5b)^2 - 15b + 9 }\)
\(\large\color{slate}{ (5b)^2 - 15b + (3)^2 }\)

so for this to be a perfect square, you must be having:
(look at the blue again)
\(\large\color{slate}{ (5b)^2 - 15b + (3)^2~~~\Rightarrow~~\color{blue}{ (5b-3)^2} }\)
Now, check if this (last blue expression (5b-3)^2 obtains the middle term.



For D,

\(\large\color{slate}{ 16x^2 - 24x + 9 }\)
\(\large\color{slate}{ (4x)^2 - 24x + 9 }\)
\(\large\color{slate}{ (4x)^2 - 24x + (3)^2 }\)

so for this to be a perfect square, you must be having:
(look at the blue again)
\(\large\color{slate}{ (4x)^2 - 24x + (3)^2~~~\Rightarrow~~\color{blue}{ (4x-3)^2} }\)
Now, check if this (last blue expression (4x-3)^2 obtains the middle term.

Answer 15

I still think C

Answer 16

AGAIN, \(\large\color{blue}{ (A-B) ^2=A^2-2AB+B^2}\)

So for your middle term to be 15, the following has to be true.
\(\large\color{slate}{ (5)(3)(2)=15~? }\)

Answer 17

see the fault here ?

Answer 18

oh.. i do. So the correct answer is D! thank you!

Answer 19

yes, you could have reduced to D using elimination

Answer 20

I will elaborater

Answer 21

\(\large\color{slate}{ (a-b)^2=a^2-2ab+b^2 }\) (we got that piece)

So if \(\large\color{slate}{ a }\) and \(\large\color{slate}{ b }\) are integers (as we had in all cases)
then, will the red part (i.e the middle term)
\(\large\color{slate}{ (a-b)^2=a^2-\color{red}{2ab}+b^2 }\)
be ever an odd number ?

Answer 22

integers, btw are
\(\large\color{slate}{ {\bf...}~-3,~-2,~-1,~0,~~1,~~2,~~3~{\bf...} }\)

all those positive and negative "whole" numbers.

Answer 23

trouble(s), concern(s), question(s) ?