Factor fully $98g^2+112g+32$ by decomposition

Question

By looking at this question I understand it is a complex trinomial so do I just decompose it??I have multiplied 98 by 32 getting 3136, but I'm not quite sure what comes next.

Answer 1

Hint: $98g^2 + 112g + 32 = (98g^2 + 56g) + (56g + 32)$. Can you take it from here?

Answer 2

$$98g^2+112g+32=2(49g^2+56g+16)=2((7g)^2+8(7g)+16)$$ Then which numbers added give you 8 and multiplied 16?

Answer 3

Note: $98g^2 + 112g + 32 = 2(49g^2 + 56g + 16) = 2(7g + 4)(7g + 4) = 2(7g + 4)^2$.

Answer 4

You already took the first step: $98 \times 32=3136$. Now, we have to find two factors of $3136$ that add up to be $112$. After some time trying out some factors, I found out that two factors of $3136$ that add up to $112$ are $56$ and $56$. Now we can split up our expression. $$98g^2+112g+32=98g^2+56g+56g+32$$ $$=14g(7g+4)+8(7g+4)$$ $$=(14g+8)(7g+4)$$ Now, we can factor out a common factor $2$ in $14g+8$. $$2(7g+4)(7g+4)$$ $$2(7g+2)^2$$ $$\displaystyle \boxed{98g^2+112g+32=2(7g+2)^2}$$

Answer 5

$98g^2 + 112 g + 32 \\= 2 \times (49 g^2 + 56g + 16) \\ = 2 \times (7^2 g^2 + 2 \times (7\times 4) g + 4^2) \\ = 2 \times (7g+4)^2$