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3 years ago

Calculate the factorization 16w^2+48w+36=(4w+[1])^2

Mathematics

1 answer:

stiv31 [10]3 years ago

3 0

Answer:

[1] = 6

x = -3/2 is the root

Step-by-step explanation:

A perfect square trinomial is of the form ...

(a + b)² = a² +2ab +b²

You have ...

a²=16w² ⇒ a = 4w
b² = 36 ⇒ b = 6
2ab = 2(4w)(6) = 48w

Then the factorization is ...

16w² +48w +36 = (4w +6)²

This will be zero when x = -6/4 = -3/2.

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Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

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Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$