BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.3, Problem 123E
To determine

To calculate: The factor of the expression (a2+1)27(a2+1)+10 .

Expert Solution

Answer to Problem 123E

The factor of the expression (a2+1)27(a2+1)+10 is (a+1)(a1)(a+2)(a2) .

Explanation of Solution

Given information:

The expression (a2+1)27(a2+1)+10 .

Formula used:

To find the factor of the trinomial of the form x2+bx+c , find two numbers r and s such that sum of the numbers is equal to coefficient of x (r+s=b) and product of two numbers is equal to constant term (rs=c) , such that (x+r) and (x+s) are the factors of x2+bx+c .

  x2+bx+c=x2+(r+s)x+rs=x2+rx+sx+rs=x(x+r)+s(x+r)=(x+r)(x+s)

The special factoring formula for difference of squares, which is mathematically expressed as,

  X2Y2=(X+Y)(XY)

Calculation:

Consider the given expression (a2+1)27(a2+1)+10 .

Recall that to find the factor of the trinomial of the form x2+bx+c , find two numbers r and s such that sum of the numbers is equal to coefficient of x (r+s=b) and product of two numbers is equal to constant term (rs=c) , such that (x+r) and (x+s) are the factors of x2+bx+c .

  x2+bx+c=x2+(r+s)x+rs=x2+rx+sx+rs=x(x+r)+s(x+r)=(x+r)(x+s)

So, find two numbers r and s such that r+s=7 and rs=10 .

Here, r=5 and s=2 , so, (a2+1)27(a2+1)+10 will be factorized as,

  (a2+1)27(a2+1)+10=(a2+1)2+(52)(a2+1)+10=(a2+1)25(a2+1)2(a2+1)+10=(a2+1)(a2+15)2(a2+15)=(a2+12)(a2+15)

Simplify the expression as,

  (a2+1)27(a2+1)+10=(a2+12)(a2+15)=(a21)(a24)=[(a)2(1)2][(a)2(2)2]

Recall the special factoring formula for difference of squares, which is mathematically expressed as,

  X2Y2=(X+Y)(XY)

Apply it,

  (a2+1)27(a2+1)+10=[(a)2(1)2][(a)2(2)2]=(a+1)(a1)(a+2)(a2)

Thus, the factor of expression (a2+1)27(a2+1)+10 is (a+1)(a1)(a+2)(a2) .

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