Question

factorization. (2) 10ײ+6×-28​

Answer 1

$\longrightarrow{\green{}}$ $10 {x}^{2} + 6x - 28$

Taking $2$ as common factor, we have

$\longrightarrow{\green{}}$ $2(5 {x}^{2} + 3x - 14)$

$\longrightarrow{\green{}}$ $2(5 {x}^{2} + 10x - 7x - 14)$

Take $5x$ as common from first two terms and $7$ from last two terms.

$\longrightarrow{\green{}}$ $2[5x(x + 2) - 7(x + 2)]$

$\longrightarrow{\green{}}$ $2(5x - 7)(x + 2)$

Answer 2

Answer:

$10x {}^{2} + 6x - 28 \\ 2(5x { }^{2} + 3x - 14) = 0 \\ 5x { }^{2} + 3x - 14 = 0 \\ 5x {}^{2} + 10x - 7x - 14 = 0 \\ 5x(x + 2) - 7(x + 2) = 0 \\ x = - 2...x = 7 \div 5$

so x= -2 and x=7/5