Identities don't have degree

Algebra Level 3

4 ( x 2 + 2 x + 1 ) ( x 2 + 3 x − 2 ) + ( x − 3 ) 2 = ( a x 2 + b x + c ) 2 4(x^2 + 2x + 1)(x^2 + 3x -2) + (x - 3)^2 \\ = (ax^2 + bx + c)^2 If the above equation is always correct, then what is the value of a 2 + b 2 + c 2 a^2 + b^2 +c^2 ?


The answer is 30.

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Vivek Parimi by Vivek Parimi , 22, India

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5 solutions

Joel Tan
Dec 9, 2014

Let a = x 2 + 2 x + 1 , b = x − 3 a=x^{2}+2x+1, b=x-3

Then the left hand side is 4 a ( a + b ) + b 2 = 4 a 2 + 4 a b + b 2 = ( 2 a + b ) 2 4a (a+b)+b^{2}=4a^{2}+4ab+b^{2}=(2a+b)^{2}

Substituting back a , b a, b the right side is ( 2 x 2 + 5 x − 1 ) 2 (2x^{2}+5x-1)^{2} .

115 Helpful 11 Interesting 17 Brilliant 15 Confused

Sadly I didn't know about this method... and expanded the whole expression out. By hand. Still got it correct, however.

Aloysius Ng - 6 years, 6 months ago

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was going to do that..but gave up because thats not how it should be done

Ashok Banerjee - 5 years, 4 months ago

can't we do it by substituting some value for x?

Charuka Bandara - 5 years, 3 months ago

Can't we substitute the values of x as 0,1,-1(say) ?? Then we get the answer quickly... ;-)

Venkata Nikhil - 5 years, 2 months ago

I’ve learned that method, but I’ve really never understood it, and you helped me understand it a little bit better thank you

Ed Jawor - 10 months, 3 weeks ago

Why can't we substitute some value for x? Why is it wrong?

Abhay Kishor Tripathi - 10 months, 1 week ago
Michael Fischer
Dec 11, 2014

Two polynomials of the same degree are equivalent iff their corresponding coefficients are equal.

Since we are given that the two fourth degree equations are equivalent, we will equate corresponding coefficients of the x^4, x^3 and the constant terms.

4 x 4 + 20 x 3 + . . . + 1 = a 2 x 4 + 2 a b x 3 + . . . + c 2 4x^4 + 20x^3+ ... +1=a^2x^4+2abx^3+ ... +c^2

So:

a 2 = 4 a^2=4

c 2 = 1 c^2=1

20 = 2 a b ⟹ 400 = 4 a 2 b 2 ⟹ b 2 = 25 20=2ab \Longrightarrow 400=4 a^2 b^2 \Longrightarrow b^2=25

a 2 + b 2 + c 2 = 30 \boxed{a^2+b^2+c^2 = 30}

44 Helpful 1 Interesting 1 Brilliant 0 Confused

Expanding L.H.S and R.H.S,we get: 4 x 4 + 20 x 3 + 21 x 2 − 10 x + 1 = a 2 x 4 + 2 a b x 3 + ( b + 2 a c ) x + 2 b c x + c 2 \color{#D61F06}{4x^4+20x^3+21x^2-10x+1=a^2x^4+2abx^3+(b+2ac)x+2bcx+c^2} Comparing coefficients,we get; a 2 = 4 → a = 2 2 a b = 20 → b = 20 2 × a = 20 2 × 2 = 5 c 2 = 1 → c = 1 a 2 + b 2 + c 2 = 2 2 + 5 2 + 1 2 = 4 + 25 + 1 = 30 \color{#3D99F6}{\begin{aligned}a^2=4\rightarrow a=2\\2ab=20\rightarrow b=\frac{20}{2\times a}=\frac{20}{2\times2}=5\\c^2=1\rightarrow c=1\\a^2+b^2+c^2=2^2+5^2+1^2=4+25+1=\boxed{30}\end{aligned}}

Abdur Rehman Zahid - 6 years, 6 months ago

I did it exactly like Joel's method but i think yours was faster. (thumbs up)

Anoorag Nayak - 5 years, 7 months ago

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Can't we compare b 2 x 2 b^2x^2 to the x 2 x^2 co-efficient ? It''s coming wrong.

Vishal Yadav - 5 years, 6 months ago

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The answer shall come ...but only when you add 2acx^2 to that ..because an x^2 coefficient is coming from that term as well.. Although u will have to take opposite signs for a and c.. i.e. a=-2 and c=1 or a=2 and c=-1 ... So this method is a bit inconvenient .. As u get multiple answers if these things are not kept in mind.. So u should proceed using only squares of a,b and c .rather than using a=-2 &+2 ,b=-5&+5 and c=+1 &-1... I hope this solves your doubt.

Rohan Kapoor - 5 years, 4 months ago

There are 2 solutions to this problem! 22 and 30. Just compute the coefficient of x² to convince yourself. Disclaimer: This is not just for you, this is for everyone who are trying to prove the answer as 30 and especially to the author of this problem!

Snehil Jain - 5 years, 6 months ago

( x ( 2 x + 5 ) − 1 ) 2 = ( x ( a x + b ) + c ) 2 (x (2 x+5)-1)^2 = (x (a x+b)+c)^2
( 2 x 2 + 5 x − 1 ) 2 = ( x ( a x + b ) + c ) 2 (2 x^2+5 x-1)^2 = (x (a x+b)+c)^2
( 2 x 2 + 5 x − 1 ) 2 = ( a x 2 + b x + c ) 2 (2 x^2+5 x-1)^2 = (a x^2+b x+c)^2

or 4 x 4 + 20 x 3 + 21 x 2 − 10 x + 1 = a 2 x 4 + 2 a b x 3 + 2 a c x 2 + b 2 x 2 + 2 b c x + c 2 4 x^4+20 x^3+21 x^2-10 x+1 = a^2 x^4+2 a b x^3+2 a c x^2+b^2 x^2+2 b c x+c^2

so, a 2 = 4 , b 2 = 25 a^2=4, b^2=25 , and c 2 = 1 c^2=1

a 2 + b 2 + c 2 = 4 + 25 + 1 = 30 a^2+b^2+c^2=4+25+1=30

6 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you arrive at the first step so quickly? That factorization doesn't seem obvious to me.

Calvin Lin Staff - 6 years, 6 months ago

FYI. To type in Latex, you just need to add brackets around your Latex code. I've edited the first part, so you can use that as a reference.

Calvin Lin Staff - 6 years, 6 months ago
Gandoff Tan
Apr 23, 2020

= 4 ( x 2 + 2 x + 1 ) ( x 2 + 3 x − 2 ) + ( x − 3 ) 2 = 4 ( x 4 + 3 x 3 − 2 x 2 + 2 x 3 + 6 x 2 − 4 x + x 2 + 3 x − 2 ) + x 2 − 6 x + 9 = 4 ( x 4 + 5 x 3 + 5 x 2 − x − 2 ) + x 2 − 6 x + 9 = 4 x 4 + 20 x 3 + 20 x 2 − 4 x − 8 + x 2 − 6 x + 9 = 4 x 4 + 20 x 3 + 21 x 2 − 10 x + 1 \begin{aligned} &\phantom{=}4\left(x^2+2x+1\right)\left(x^2+3x-2\right)+(x-3)^2\\ &=4\left(x^4+3x^3-2x^2+2x^3+6x^2-4x+x^2+3x-2\right)+x^2-6x+9\\ &=4\left(x^4+5x^3+5x^2-x-2\right)+x^2-6x+9\\ &=4x^4+20x^3+20x^2-4x-8+x^2-6x+9\\ &=4x^4+20x^3+21x^2-10x+1 \end{aligned}

= ( a x 2 + b x + c ) 2 = a 2 x 4 + a b x 3 + a c x 2 + a b x 3 + b 2 x 2 + b c x + a c x 2 + b c x + c 2 = a 2 x 4 + 2 a b x 3 + ( b 2 + 2 a c ) x 2 + 2 b c x + c 2 \begin{aligned} &\phantom{=}\left(ax^2+bx+c\right)^2\\ &=a^2x^4+abx^3+acx^2+abx^3+b^2x^2+bcx+acx^2+bcx+c^2\\ &=a^2x^4+2abx^3+\left(b^2+2ac\right)x^2+2bcx+c^2 \end{aligned}

4 ( x 2 + 2 x + 1 ) ( x 2 + 3 x − 2 ) + ( x − 3 ) 2 = ( a x 2 + b x + c ) 2 4 x 4 + 20 x 3 + 21 x 2 − 10 x + 1 = a 2 x 4 + 2 a b x 3 + ( b 2 + 2 a c ) x 2 + 2 b c x + c 2 \begin{aligned} 4\left(x^2+2x+1\right)\left(x^2+3x-2\right)+(x-3)^2&=\left(ax^2+bx+c\right)^2\\ 4x^4+20x^3+21x^2-10x+1&=a^2x^4+2abx^3+\left(b^2+2ac\right)x^2+2bcx+c^2 \end{aligned}

{ a 2 = 4 2 a b = 10 c 2 = 1 ⟹ { a 2 = 4 a b = 10 c 2 = 1 ⟹ { a 2 = 4 a 2 b 2 = 100 c 2 = 1 ⟹ { a 2 = 4 4 b 2 = 100 c 2 = 1 ⟹ { a 2 = 4 b 2 = 25 c 2 = 1 \begin{cases} a^2=4\\ 2ab=10\\ c^2=1 \end{cases} \implies \begin{cases} a^2=4\\ ab=10\\ c^2=1 \end{cases} \implies \begin{cases} a^2=4\\ a^2b^2=100\\ c^2=1 \end{cases} \implies \begin{cases} a^2=4\\ 4b^2=100\\ c^2=1 \end{cases} \implies \begin{cases} a^2=4\\ b^2=25\\ c^2=1 \end{cases}

a 2 + b 2 + c 2 = 4 + 25 + 1 = 30 a^2+b^2+c^2=4+25+1=\boxed{30}

3 Helpful 1 Interesting 2 Brilliant 0 Confused
Jason Hughes
Jan 8, 2015

First plug in x=0 to get c^2=1. Notice the x^4 terms on both sides are 4x^4=a^2x^4. so a^2=4. Then plug in x=-1 to get 16=(a-b+c)^2. Plug in x=1 to get 36=(a+b+c)^2 add the two to get 52=2(a^2+b^2+c^2+2ac). Simplify to get 26=a^2+b^2+c^2+2ac. Plug in x=-2 to get 9=(4a-2b+c)^2. Plug in x=2 to get 289=(4a+2b+c)^2. Add those two get 298=2(16a^2+4b^2+c^2+8ac). Simplify and plug in values of a^2 and c^2 to get. b^2+2ac=21. We know a^2*c^2=4 so ac is either 2 or -2. for (ax^2+bx+c)^2 to expand to integer coefficients a,b, c must be whole integers so ac=-2. so b^2=25. Then a^2+b^2+c^2=26+2 * 2= 30.

3 Helpful 0 Interesting 1 Brilliant 1 Confused

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