Question

factorise using identity square 2xsquare -32x+78​

Answer 1

$Given \: Quadratic \: expression : \\2x^{2} - 32 x + 78$

$= 2( x^{2} - 16x+ 39 )$

$= 2[ x^{2} + ( -13 -3 )x + (-13)(-3)]$

$\underline { \blue { By \: Algebraic \: identity:}}$

$\boxed { \pink { x^{2} + (m+n)x + mn = (x+m)(x+n) }}$

$= 2(x-13)(x-3)$

Therefore.,

$\red { 2x^{2} - 32 x + 78 } \green { = 2(x-13)(x-3)}$

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Answer 2

hi mate,

solution: factorise using identity

2x² -32x+78

Pull out like factors :

2x² - 32x + 78 = 2 • (x² - 16x + 39)

Factoring x² - 16x + 39

x² - 13x - 3x - 39

Step-1 : Add up the first 2 terms, pulling out like factors :

x • (x-13)

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

3 • (x-13)

Step-2 : Add up the four terms of step 1 :

(x-3) • (x-13)

Which is the desired factorization

factorise by using identity

(x +a) (x + b) = x(x + b) + a(x + b)

Hence, we can write = x2 + bx +ax + ab

Or (x + a) (x + b) = x2 + (a +b)x +ab

Final result :

2 • (x - 3) • (x - 13)

i hope it helps you..