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Airida [17]
3 years ago
9

Find the missing factor. 4b2 + 17b + 15 = (b + 3)( )

Mathematics
1 answer:
kari74 [83]3 years ago
5 0

Answer:

(4b + 5)

Step-by-step explanation:

To get 4b^2 you already have b^2 if you put b  inside the second set of brackets. But that would mean you don't have 4 anywhere to get 4b^2.

So the first step has to be

(b + 3)(4b

Now look at the 15 for a moment. It is plus 15. The only way you can get a plus 15 is if both signs are plus (after the b terms) or both terms are minus.

The middle term (17b) is plus so both terms after b are plus.

(b + 3)(4b +

Now we need something that multiplies to 15. 3*5 = 15. So the term you want is 5.

(b + 3)(4b + 5)

Does the middle term work?

5*b + 3*4b = 5b + 12b = 17b

Everything looks fine.

The second factor is 17b.

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1^o\\5\leq10\leq x\\\\\left\{\begin{array}{ccc}5+10 \ \textgreater \  x\\5^2+10^2 \ \textgreater \  x^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x^2 \ \textless \  125\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \  15\\ x \ \textless \  \sqrt{125}\approx11.1\end{array}\right\\\\\boxed{x=11}\\\\2^o\\5\leq x\leq10\\\\\left\{\begin{array}{ccc}5+x \ \textgreater \  10\\ 5^2+x^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x^2 \ \textgreater \  75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \  5\\ x \ \textgreater \  \sqrt{75}\approx8.7\end{array}\right\\\boxed{x=9}

3^o\\x\leq5\leq10\\\\\left\{\begin{array}{ccc}x +5 \ \textgreater \  10\\ x^2+5^2 \ \textgreater \  10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x\in\O}\\\\Answer:\boxed{x=9\ or\ x=11}\to your\ answer:\boxed{\boxed{x=11}}
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