Question

FACTORISE:
P^2+ 4p + 3

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Answer 1

★ Concept:-

Here this is a question about factorisation. Factorisation actually means finding of roots or making the given equation into its factors.

We have 2 methods for factorisation:

$\sf\bullet First \longrightarrow Middle\: term\: splitting.$

$\sf \bullet Second\longrightarrow Quadratic\: formula.$

So let's solve it by both methods.

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$\huge\underline{\boxed{\sf{METHOD\; 1 }}}\bigstar$

 

★ Middle term splitting:

$:\longrightarrow$  P²+4P+3=0

$:\longrightarrow$  P²+3P+P+3=0

$:\longrightarrow$  P(P+3)+1(P+3)=0

$:\longrightarrow$ (P+1)(P+3)=0

$:\longrightarrow$  P=-1,-3

∵ Roots are -1 and -3.

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$\huge\underline{\boxed{\sf{METHOD\; 2 }}}\bigstar$

★ Quadratic  

In the given equation we have:-

$:\longrightarrow$ Coefficient of P², a=1

$:\longrightarrow$ Coefficient of P, b=4

$:\longrightarrow$ Constant term , c=3

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~Now apply formula:-

$\large\underline{\boxed{\sf X=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}}\bigstar$

Now putting the values:

$\sf: \longrightarrow X=\dfrac{-4\pm\sqrt{4^2-4(1)(3)}}{2(1)}$

$\sf: \longrightarrow X=\dfrac{-4\pm\sqrt{16-12}}{2}$

$\sf: \longrightarrow X=\dfrac{-4\pm\sqrt{4}}{2}$

$\sf: \longrightarrow X=\dfrac{-4\pm2}{2}$

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For positive:-

$\sf: \longrightarrow X=\dfrac{-4+2}{2}$

$\sf: \longrightarrow X=\dfrac{-2}{2}$

$\sf: \longrightarrow X=\dfrac{-\not 2}{\not 2}$

$\Large\boxed{\sf{\red{\implies X=-1}}}$

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For negative:-

$\sf: \longrightarrow X=\dfrac{-4-2}{2}$

$\sf: \longrightarrow X=\dfrac{-6}{2}$

$\sf: \longrightarrow X=\dfrac{-\not 6}{\not 2}$

$\Large\boxed{ \sf{\pink{ \implies X=-3}}}$

∵ The value of roots are -1 and -3.

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