Expanding Double and Triple Brackets

Multiply each term in the first bracket by each term in the second.

1. Multiply each term in the first bracket by each term in the second: x\times x=x^2, x\times4=4x, 3\times x=3x and 3\times4=12.
2. So (x+3)(x+4)=x^2+4x+3x+12.
3. Collect the like terms 4x+3x=7x: the answer is x^2+7x+12.

Keep the negative sign when multiplying by the second term.

1. Multiplying out gives x^2-2x+6x-12.
2. Collect like terms to obtain x^2+4x-12.

There are four products before you collect like terms.

1. Multiplying out gives 3x^2+15x+2x+10.
2. The linear terms add to 17x, so the result is 3x^2+17x+10.

Multiply the negative term by both terms in the other bracket.

1. The products give 6x^2+2x-9x-3.
2. Collecting like terms gives 6x^2-7x-3.

A squared bracket means the bracket multiplied by itself.

1. (x-4)^2 means (x-4)(x-4), not x^2-4^2.
2. The four products are x\times x=x^2, x\times(-4)=-4x, -4\times x=-4x and -4\times(-4)=+16.
3. Collect the two -4x terms: x^2-8x+16.

Look for two linear terms that cancel.

1. Multiplying out gives 25x^2-15x+15x-9.
2. The middle terms cancel, leaving 25x^2-9.

Treat each letter as part of its term and collect the terms containing both letters.

1. The four products are 2a^2, 8ab, -3ab and -12b^2.
2. Combining the mixed terms gives 2a^2+5ab-12b^2.

Only the two cross products contribute to the linear term.

1. x\times3=3x and -5\times2x=-10x.
2. Their sum is -7x, so the coefficient is -7.

Expand both products before subtracting the whole of the second result.

1. The products are x^2+3x-4 and x^2-x-6.
2. Subtracting gives x^2+3x-4-x^2+x+6.
3. Collect like terms: 4x+2.

Expand two brackets first, then multiply every term by the remaining bracket.

1. Expand two of the brackets first: (x+1)(x+2)=x^2+2x+x+2=x^2+3x+2.
2. Now multiply each term of x^2+3x+2 by x and then by 4.
3. Multiplying by x gives x^3+3x^2+2x; multiplying by 4 gives 4x^2+12x+8.
4. Collect like terms: x^3+(3+4)x^2+(2+12)x+8=x^3+7x^2+14x+8.

Write the square as two identical brackets and look at both cross products.

1. (2x+3)^2 means (2x+3)(2x+3), so there are four products: 4x^2, 6x, 6x and 9.
2. Each cross product is 2x\times3=6x, and there are two of them, so together they give 12x.
3. The correct expansion is 4x^2+12x+9, so the missing term is 12x.
4. Choosing 6x counts only one of the two cross products.

Multiply the number of rows by the number of seats in each row.

1. The total is (n+2)(3n-1).
2. Expansion gives 3n^2-n+6n-2.
3. So the total number of seats is 3n^2+5n-2.

Apply the factor outside the product after expanding its two brackets.

1. The square expands to x^2-6x+9.
2. The other term is 2(x^2-x-2)=2x^2-2x-4.
3. Adding and collecting gives 3x^2-8x+5.

Expand two brackets first, then collect terms with the same power after multiplying by the third.

Part (a)

1. First, (x-2)(x+3)=x^2+x-6.
2. When this is multiplied by (x+5), the squared terms are 5x^2 and x^2.
3. They add to 6x^2, so the coefficient is 6.

Part (b)

1. The linear terms in (x^2+x-6)(x+5) are 5x and -6x.
2. Their sum is -x, so the coefficient is -1.

Expand both products; the squared terms will cancel.

1. Expand each product: (x+5)(x-1)=x^2+4x-5 and (x+1)(x-3)=x^2-2x-3.
2. Subtract the whole of the second expansion, keeping it in a bracket: (x^2+4x-5)-(x^2-2x-3)=x^2+4x-5-x^2+2x+3.
3. The x^2 terms cancel, leaving 6x-2. So 6x-2=28.
4. Add 2: 6x=30, so x=5.

Start with either pair of brackets, keeping careful track of negative terms.

1. Expand two brackets first: (x-4)(3x+2)=3x^2+2x-12x-8=3x^2-10x-8.
2. Multiply each term of 3x^2-10x-8 by 2x: 6x^3-20x^2-16x.
3. Multiply each term of 3x^2-10x-8 by 1: 3x^2-10x-8.
4. Add and collect like terms: 6x^3+(-20+3)x^2+(-16-10)x-8=6x^3-17x^2-26x-8.

Choose the first pair so that their cross products cancel.

1. Start with (x-3)(x+3): the cross products 3x and -3x cancel, so it is a difference of two squares, x^2-9.
2. Now multiply each term of x^2-9 by (2x-5): x^2\times2x=2x^3, x^2\times(-5)=-5x^2, -9\times2x=-18x and -9\times(-5)=45.
3. There are no like terms to collect, so the answer is 2x^3-5x^2-18x+45.

Expand the brackets without the unknown constant first.

1. Expand the two brackets without k first: (x-2)(x+4)=x^2+2x-8.
2. Now multiply (x^2+2x-8)(x+k). The x^2 terms come from x^2\times k=kx^2 and 2x\times x=2x^2.
3. So the coefficient of x^2 is k+2.
4. Set k+2=9, so k=7.

Expand each product and remember to subtract every term in the second product.

1. First product: (x+2)(x+3)=x^2+5x+6. Multiply by (x-1): x^3+5x^2+6x-x^2-5x-6=x^3+4x^2+x-6.
2. Second product: (x+1)(x+3)=x^2+4x+3. Multiply by x: x^3+4x^2+3x.
3. Subtract the whole of the second product, keeping it in a bracket: (x^3+4x^2+x-6)-(x^3+4x^2+3x).
4. The x^3 and x^2 terms cancel, and x-3x=-2x, so the answer is -2x-6.

Expand the two known brackets and compare coefficients of matching powers.

Part (a)

1. Expand the two known brackets: (2x-1)(x+3)=2x^2+6x-x-3=2x^2+5x-3.
2. Multiplying (2x^2+5x-3)(x+p), the only number term is -3\times p=-3p.
3. The two sides are equal for all x, so their number terms are equal: -3p=-12.
4. So p=4.

Part (b)

1. In (2x^2+5x-3)(x+p), the x terms come from 5x\times p=5px and -3\times x=-3x.
2. So the coefficient of x is 5p-3, which must equal q.
3. Using p=4: q=5\times4-3=17.
4. Check with the x^2 terms: 2p+5=2\times4+5=13, which matches 13x^2.