Factorise 3x^2+13x+4 by splitting the middle term

Mathematics

1 answer:

mihalych1998 [28]3 years ago

4 0

Answer:

( x + 4 ) ( 3x + 1 )

Step-by-step explanation:

→ Identify the product, sum & factors

Product = 4

Sum = 13

Factors = 4 & 1

3x² + 12x + x + 4

→ Factorise 3x² + 12x and x + 4

3x² + 12x ⇔ 3x ( x + 4 )

x + 4 ⇔ 1 ( x + 4 )

→ Take out the ( x + 4 ) as a factor and ( 3x + 1 )

( x + 4 ) ( 3x + 1 )

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Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ , if f(x) approaches L as x approaches 'a'.

(a)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\$

$\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

(b)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$