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Question: The value of the expression \(\dfrac{{{x}^{2}}+7x+12}{x+3}\) [a] x+4 [b] x+7 [c] x [d] None...

The value of the expression x2+7x+12x+3\dfrac{{{x}^{2}}+7x+12}{x+3}
[a] x+4
[b] x+7
[c] x
[d] None

Explanation

Solution

Hint: Factorise the above numerator of the above expression by using splitting the middle term method or completing the square method. Cancel out the common factors between numerator and denominator. Alternatively, you can use a long division method or synthetic division method to find the value of the above expression.

Complete step-by-step answer:

First, we will check that if x=-3 is a factor of the numerator. For that, we will use the factor theorem.
Let P(x) be a polynomial in x, then x-a is a factor of P(x) if and only if P(a) = 0
Here P(x) =x2+7x+12={{x}^{2}}+7x+12 and a = -3
Now P(-3) =(−3)2+7(−3)+12=9−21+12=0={{\left( -3 \right)}^{2}}+7\left( -3 \right)+12=9-21+12=0
Since P(-3) = 0, x+3 is a factor of P(x)
Hence the above expression is reducible
Let P(x)= (x+a)(x+3) as x+3 is a factor of p(x)
So we have x2+7x+12=x2+(3+a)x+3a{{x}^{2}}+7x+12={{x}^{2}}+\left( 3+a \right)x+3a
Comparing constant terms we get
3a = 12
i.e. a = 4
Hence P(x) = (x+3)(x+4)
Hence
P(x)x+3=x+4 ⇒x2+7x+12x+3=x+4 \begin{aligned} & \dfrac{P(x)}{x+3}=x+4 \\\ & \Rightarrow \dfrac{{{x}^{2}}+7x+12}{x+3}=x+4 \\\ \end{aligned}
Hence option [a] is correct.

Note: [1] Alternatively we have
x2+3x=x(x+3){{x}^{2}}+3x=x\left( x+3 \right)
Adding 4x on both sides x2+7x=x(x+3)+4x{{x}^{2}}+7x=x\left( x+3 \right)+4x
Adding 12 on both sides, we get

& {{x}^{2}}+7x+12=x\left( x+3 \right)+4x+12 \\\ & =x\left( x+3 \right)+4\left( x+3 \right) \\\ & =\left( x+3 \right)\left( x+4 \right) \\\ \end{aligned}$$ [2] Using Synthetic division $\begin{aligned} & {{x}^{2}}+7x+12 \\\ & -3|\ \ \ \ 1\text{ 7 12} \\\ & \text{ 0 -3 -12} \\\ & \text{ 1 4 0} \\\ \end{aligned}$ Hence $\dfrac{{{x}^{2}}+7x+3}{x+3}=x+4$ [3] Using the long division method Leading term in numerator i.e ${{x}^{2}}$ can be cancelled by multiplying x to x+3 and subtracting it from numerator We get q = x and r $={{x}^{2}}+7x+12-{{x}^{2}}-3x=4x+12$ Leading term in r can be cancelled by multiplying 4 to x+3 and subtracting it from r. We get q = x + 3 and r = 4x+12-4(x+3) = 4x+12-4x-12=0 Hence $\dfrac{{{x}^{2}}+7x+12}{x+3}=x+4$