Which is the decomposed form for the expression x^2 + 4x + 3?

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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It is required to decompose x^2+4x+3.

We shal treat the given expression of secon degree to consist of 2 factors like x+a and x+b. Then the product of the x+a and x+b must be the given expression. So.

(x+a)(x+b) = x^2+4x+3. So,

X^2+(a+b)x+ab = x^2+4x+3. Comparing the coefficients of x^2, x and constant terms on both sides, we get:

1/1 = 4/(a+b) = 3/(ab). Therefore,

a+b =4 and

ab =3.

Solving these inequality, you get, (We do it by guess here without prolonging the procedure): a = 3 and b =1.

Threfore, x^2+4x+3 = (x+3)(x+1)

 

 

 

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

I believe, by decomposed for of the expression you mean a form in which the expression is represented as multiple of two or more constituent factors.

The given expression is:

E = x^2 + 4x + 3

We can find its factors by suitably modifying the expression in stages without changing its value. For the given expression it can be done as follows:

E = x^2 + 4x + 3

= x^2 + 3x + x + 3 = (x^2 + 3x) + (x + 3)

= x*(x + 3) + 1*(x + 3)

= (x + 1)*(x + 3)

Thus factors of the given expression are (x + 1) and (x + 3).

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We’ll consider the expression as a second grade polynomial.  

A polynomial could be written as a product of linear factors, according to his roots. For this reason, first of all, we’ll try to find out the roots of the expression.

x^2+4x+3=0

 The coefficients of the polynomial x^2+4x+3=0 are a=1, b=4, c=3

x1=[-b + sqrt (b^2-4*a*c)]/2*a

x1=[-4 + sqrt (16-4*1*3)]/2*1

x1= (-4+sqrt4)/2

x1=(-4+2)/2

 x1=-1

From Viete first relation between roots and coefficients, we’ll have:

x1+x2=-(b/a)

-1 + x2=-4 => x2=-4+1

x2=-3

After finding the roots, we could write the expression as a product of linear factors.

x^2+4x+3= (x-x1)*(x-x2)

x^2+4x+3= (x+1)*(x+3)

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