Question

Factorise the following bye splitting the middle term a)2b^2+32b+128

Answer 1

Answer:

STEP

1

:

Equation at the end of step 1

(2b2 - 32b) + 128

STEP

2

:

STEP

3

:

Pulling out like terms

3.1 Pull out like factors :

2b2 - 32b + 128 = 2 • (b2 - 16b + 64)

Trying to factor by splitting the middle term

3.2 Factoring b2 - 16b + 64

The first term is, b2 its coefficient is 1 .

The middle term is, -16b its coefficient is -16 .

The last term, "the constant", is +64

Step-1 : Multiply the coefficient of the first term by the constant 1 • 64 = 64

Step-2 : Find two factors of 64 whose sum equals the coefficient of the middle term, which is -16 .

-64 + -1 = -65

-32 + -2 = -34

-16 + -4 = -20

-8 + -8 = -16 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -8 and -8

b2 - 8b - 8b - 64

Step-4 : Add up the first 2 terms, pulling out like factors :

b • (b-8)

Add up the last 2 terms, pulling out common factors :

8 • (b-8)

Step-5 : Add up the four terms of step 4 :

(b-8) • (b-8)

Which is the desired factorization

Answer 2

$\huge { \boxed {\bf{question}}}$

factorise the following by splitting the middle term ;

$a) \: {2b}^{2} + 32b + 128$

$\huge { \boxed {\bf{answer}}}$

How to find it ?

=> By Using The Splitting Method

Now ,

according to given question ,

${2b}^{2} + 32b + 128$

$\rightarrow \: by \: factor \: 2\: out \: of \: the \: expression$

$2( {b}^{2} + 16b + 64)$

$\rightarrow \: now \: by \: using \: {a}^{2} + 2ab + {b}^{2} = (a + b) ^{2}$

$2(b + 8) ^{2}$

So, the answer is

$2(b + 8) (b+8)$