Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Which ones are a perfect square trinomials? 9x^2− 6x + 1 36b^2-24b + 8 16x^2+ 24x + 9 4a^2− 10a + 25

OpenStudy (mathmate):

Perfect square trinomials are trinomials where 1. The first and last terms are perfect squares. 2. The coefficient of the second term equals double the product of the square-roots of the first and third terms. I will give an example: \( x^2+8x+16 \) 1. 1 and 16 are both perfect squares. The square roots of these coefficients are 1 and 4. 2. Double the product of 1 and 4 gives 8, which matches the coefficient of the second term. Therefore the given trinomial is a perfect square trinomial.

OpenStudy (mathmate):

An additional note: The middle coefficient could be positive or negative and still be a perfect square trinomial.

OpenStudy (anonymous):

But what if the coefficient is not a 1?

OpenStudy (mathmate):

\( 25x^2+40x+16\) 25 and 16 are perfects (square roots = 5 and 4) double the product =2*5*4=40 = coef. of middle term. => \( 25x^2+40x+16\) is a perfect square trinomial.

OpenStudy (mathmate):

The factorization is : \( 25x^2+40x+16\) = \( (5x+4)^2 \)

OpenStudy (mathmate):

If the middle coefficient is negative, the it would be (5x-4)^2

OpenStudy (anonymous):

So is the 1st, 2nd, and last one a perfect trinomial?

OpenStudy (mathmate):

For the second one, if you factor out the common factor, the last term is 2, which is not a perfect square, so... By now, you can factorize (by sight) the first, third and fourth.

OpenStudy (anonymous):

Ohh ok THANK YOUU @mathmate !!!!!!!!!

OpenStudy (mathmate):

You're welcome! Note: Do remember that the first and last term must be BOTH positive or negative for the trinomial to be a perfect square. If they are both negative, then factor out the negative sign, example: \( -3x^2+6x-3 \\\) can be factored as \( -3(x^2-2x+1) \) The part in parentheses is a perfect square trinomial, but with the factor, it won't be, because -3 is itself not a perfect square.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!