Identities and Comparing Coefficients

An identity is true for every value of x, so simplify each left-hand side and see whether it matches the right-hand side exactly.

1. An identity (\equiv) must be true for every value of x
2. (x-4)^2=x^2-8x+16, not x^2-16
3. 3(2x-1)=6x-3, not 6x-1
4. x(x+2)-2x=x^2+2x-2x=x^2, true for every x
5. 2(x+3)=x+8 is only true when x=2, so it is an equation, not an identity
6. Answer: x(x+2)-2x\equiv x^2

Multiply the two numbers in the brackets together and compare with the constant term on the right.

1. The constant term of (x+4)(x-k) is 4\times(-k)=-4k
2. Compare constants: -4k=-24, so k=6
3. Check the x terms: 4x-6x=-2x ✓
4. Answer: k=6

Expand the left-hand side, then compare the constant terms on each side.

1. Expand: 12x-4k+2x+10
2. Collect like terms: 14x+(10-4k)
3. The x terms already match: 14x
4. Compare the constant terms: 10-4k=-6
5. -4k=-16
6. k=4

Expand each squared bracket in full, then subtract the whole of the second expansion.

1. (x+a)^2=x^2+2ax+a^2
2. (x-a)^2=x^2-2ax+a^2
3. Subtract: x^2+2ax+a^2-x^2+2ax-a^2=4ax
4. Compare with 20x: 4a=20
5. Answer: a=5

You don't need the whole expansion: the x^2 terms come only from the first terms of each bracket, and the constant from the last terms.

Part (a)

1. The x^2 term on the left is 5x\times px=5px^2
2. Compare x^2 terms: 5p=10
3. Answer: p=2

Part (b)

1. The constant term on the left is -2\times q=-2q
2. Compare constants: -2q=-6, so q=3
3. Check the x terms: 5qx-2px=15x-4x=11x ✓
4. Answer: q=3

Expand both sides, then compare the x terms and the constant terms: q appears on both sides, so collect it on one side.

Part (a)

1. Left: 2px+6+q
2. Right: 8x+4q-9
3. Compare x terms: 2p=8
4. Answer: p=4

Part (b)

1. Compare constants: 6+q=4q-9
2. 15=3q
3. Answer: q=5

Expand as usual, treating y like a number; then compare the y^2 terms first.

Part (a)

1. Expand: 3x^2-6xy+pxy-2py^2
2. Compare the y^2 terms: -2p=-10
3. Answer: p=5

Part (b)

1. The xy terms are -6xy+pxy=(p-6)xy
2. With p=5: q=5-6
3. Answer: q=-1

Start with the constant term: what must k^3 equal?

1. The constant term of (x+k)^3 is k^3
2. So k^3=-64, which gives k=-4
3. (x-4)^2=x^2-8x+16
4. (x-4)^3=(x-4)(x^2-8x+16)=x^3-12x^2+48x-64
5. Compare the x terms: b=48

Multiply the first two brackets together first: they make a difference of two squares.

1. (x+k)(x-k)=x^2-k^2
2. (x^2-k^2)(x+3)=x^3+3x^2-k^2x-3k^2
3. Compare x terms: -k^2=-16, so k^2=16
4. Compare constants: c=-3k^2=-3\times16
5. Answer: c=-48 (whether k=4 or k=-4)

Multiply every term by 4 to clear the fractions, remembering that the minus sign applies to the whole of (x-q).

Part (a)

1. Multiply every term by 4: 2(px+3)-(x-q)\equiv 5x+13
2. Expand: 2px+6-x+q\equiv 5x+13
3. Compare x terms: 2p-1=5
4. Answer: p=3

Part (b)

1. Compare constants: 6+q=13
2. Answer: q=7

Comparing the x terms gives one equation and comparing the constants gives another: solve them simultaneously.

Part (a)

1. Expand: px+3p+2qx-q
2. Compare x terms: p+2q=11
3. Compare constants: 3p-q=12
4. From the second equation, q=3p-12
5. Substitute: p+2(3p-12)=11, so 7p=35
6. Answer: p=5

Part (b)

1. q=3p-12=15-12
2. Answer: q=3

Expand both brackets, then subtract the whole of (x+1)(x-3); compare the x^2 terms first to find k.

Part (a)

1. k(x-2)^2=kx^2-4kx+4k
2. (x+1)(x-3)=x^2-2x-3
3. Subtract: (k-1)x^2+(2-4k)x+4k+3
4. Compare x^2 terms: k-1=2, so k=3
5. Compare x terms: p=2-4k=2-12
6. Answer: p=-10

Part (b)

1. Compare constants: q=4k+3
2. q=12+3
3. Answer: q=15

You only need the terms of the expansion that give the constant, then the x terms, then the x^2 terms.

Part (a)

1. The constant term on the left is -2\times5=-10
2. Answer: r=-10

Part (b)

1. The x terms on the left are 5x-2px
2. Compare with -x: 5-2p=-1
3. Answer: p=3

Part (c)

1. The x^2 terms on the left are px^2-2x^2
2. q=p-2=3-2
3. Answer: q=1

Expand p(x-q)^2+r and compare the x^2 terms, then the x terms, then the constants.

Part (a)

1. Compare x^2 terms: p=3
2. Answer: p=3

Part (b)

1. p(x-q)^2+r=px^2-2pqx+pq^2+r
2. Compare x terms: -2pq=-9, so -6q=-9
3. Answer: q=\dfrac{3}{2}

Part (c)

1. Compare constants: pq^2+r=1
2. 3\times\dfrac{9}{4}+r=1, so \dfrac{27}{4}+r=1
3. r=1-\dfrac{27}{4}
4. Answer: r=-\dfrac{23}{4}

Expand and collect the left-hand side into x^2 terms, x terms and constants, then compare them in that order.

Part (a)

1. Expand: p(x^2+4x+4)+qx+2q+r
2. =px^2+(4p+q)x+4p+2q+r
3. Compare x^2 terms: p=3
4. Compare x terms: 4p+q=5, so 12+q=5
5. Answer: q=-7

Part (b)

1. Compare constants: 4p+2q+r=-4
2. 12-14+r=-4
3. Answer: r=-2

Compare the constant terms and the x^2 terms: they give you the product and the sum of p and q. Answer part (a) before attempting part (b).

Part (a)

1. Compare constants: 2pq=24, so pq=12
2. The x^2 terms on the left are (p+q+2)x^2
3. Compare x^2 terms: p+q+2=9, so p+q=7
4. Two numbers with sum 7 and product 12: 4 and 3
5. p>q, so p=4 and q=3
6. Answer: p=4

Part (b)

1. (x+4)(x+3)=x^2+7x+12
2. (x^2+7x+12)(x+2)=x^3+9x^2+26x+24
3. Answer: r=26

Expand the left-hand side fully and collect the x terms and the constant terms before comparing with the right-hand side.

Part (a)

1. Expand: (x+a)^2-b(x+3)=x^2+2ax+a^2-bx-3b
2. The x terms are (2a-b)x
3. Compare with 2x: 2a-b=2
4. b=2a-2

Part (b)

1. Compare the constant terms: a^2-3b=-2
2. Substitute b=2a-2 (use a bracket): a^2-3(2a-2)=-2
3. a^2-6a+6=-2
4. a^2-6a+8=0
5. (a-2)(a-4)=0
6. a=2 or a=4

Part (c)

1. a=4, so b=2(4)-2=6
2. Check: (x+4)^2-6(x+3)=x^2+8x+16-6x-18=x^2+2x-2 ✓

Expand (px+q)^2 and compare the x^2 terms, the x terms and the constants in turn: one of them gives a quadratic equation in q.

Part (a)

1. Expand: p^2x^2+2pqx+q^2
2. Compare x^2 terms: p^2=4, so p=2 or p=-2
3. Compare x terms: 2pq=4q
4. q\neq0 (the constants would give 0=15), so 2p=4
5. Answer: p=2

Part (b)

1. Compare constants: q^2=2q+15
2. q^2-2q-15=0
3. (q-5)(q+3)=0
4. Answer: q=5 or q=-3

Expand (2n+p)^2 and compare coefficients. Answer part (a) before attempting part (b).

Part (a)

1. Expand: (2n+p)^2+q=4n^2+4pn+p^2+q
2. Compare n terms: 4p=12, so p=3
3. Compare constants: p^2+q=13, so 9+q=13
4. Answer: q=4

Part (b)

1. 4n^2+12n+13=(2n+3)^2+4
2. 2n+3 is an odd integer, so it can't be 0
3. The smallest value of (2n+3)^2 is 1, when n=-1 or n=-2
4. Smallest value =1+4
5. Answer: 5

Part (c)

1. 2n+3 is odd, so (2n+3)^2 is odd
2. Odd +\,4 is odd, so (2n+3)^2+4 is always odd
3. It is not always prime: n=3 gives 85=5\times17
4. It is not always greater than 13: n=-1 gives 5
5. Answer: 4n^2+12n+13 is always odd

An identity is true for every value of x: choose a value of x that makes two of the three terms on the left equal to zero.

Part (a)

1. An identity is true for every value of x
2. Put x=1: the first two terms are 0
3. r(-1)(-2)=4-9+7, so 2r=2
4. Answer: r=1

Part (b)

1. Put x=2: the first and third terms are 0
2. q(1)(-1)=16-18+7, so -q=5
3. Answer: q=-5

Part (c)

1. Compare x^2 terms: p+q+r=4
2. p-5+1=4
3. Answer: p=8 (or put x=3: 2p=16)