Both terms divide by \(3\), so take it out first.

\(27x^2 - 192 = 3(9x^2 - 64)\)

What is left is a difference of two squares, because \(9x^2 = (3x)^2\) and \(64 = 8^2\).

Using \(A^2 - B^2 = (A + B)(A - B)\) with \(A = 3x\) and \(B = 8\),

\(9x^2 - 64 = (3x + 8)(3x - 8)\)

So altogether

\(3(3x + 8)(3x - 8)\)

Multiplying the brackets out has to give back the left-hand side, so compare a term you can see all of.

The \(v^2\) term comes from \(4v\) times \(gv\) and nothing else.

\(4g = 20\)

\(g = 20 \div 4 = 5\)

The constant term comes from \(-9\) times \(h\) and nothing else.

\((-9) \times h = -54\)

\(h = -54 \div (-9) = 6\)

Check it against the \(v\) term as well: \(4 \times 6 + 5 \times (-9) = -21\)

Multiply the coefficient of \(w^2\) by the constant term.

\(15 \times 12 = 180\)

Now find two numbers that multiply to give \(180\) and add to give \(28\). They are \(10\) and \(18\).

Use them to split the middle term.

\(15w^2 + 10w + 18w + 12\)

Take a factor out of the first two terms, then out of the last two.

\(5w(3w + 2) + 6(3w + 2)\)

Both parts now share the bracket \((3w + 2)\), so it comes out.

\((3w + 2)(5w + 6)\)

Look at the three coefficients first: \(15\), \(-48\) and \(-48\) all divide by \(3\), so take that out before anything else.

\(15y^2 - 48y - 48 = 3(5y^2 - 16y - 16)\)

Now factorise the bracket. Multiply its \(y^2\) coefficient by its constant term: \(5 \times (-16) = -80\), and look for two numbers that multiply to \(-80\) and add to \(-16\). They are \(-20\) and \(4\).

\(5y^2 - 20y + 4y - 16\)

\(5y(y - 4) + 4(y - 4)\)

\(= (5y + 4)(y - 4)\)

Putting the \(3\) back gives

\(3(5y + 4)(y - 4)\)

Factorise the left-hand side first. Multiply the coefficient of \(x^2\) by the constant term.

\(12 \times (-35) = -420\)

Two numbers multiplying to \(-420\) and adding to \(-13\) are \(15\) and \(-28\), so split the middle term with them.

\(12x^2 + 15x - 28x - 35\)

\(3x(4x + 5) - 7(4x + 5) = (4x + 5)(3x - 7)\)

A product is zero only when one of its factors is zero.

\(4x + 5 = 0\) gives \(x = -\frac{5}{4}\)

\(3x - 7 = 0\) gives \(x = \frac{7}{3}\)

Multiply the first and last numbers: \(4\) and \(-49\) give \(-196\).

Find two numbers with product \(-196\) and sum \(21\): they are \(28\) and \(-7\).

Split the middle term:

\(4n^2 + 28n - 7n - 49\)

Factorise in pairs:

\(4n(n + 7) - 7(n + 7)\)

\((4n - 7)(n + 7)\)

Use the factorised form \((4n - 7)(n + 7)\). Every term is the product of the two brackets.

The second bracket \(n + 7\) is greater than 1 for every term, so a term can only be prime if the first bracket is 1. A negative first bracket gives a negative term, which is not prime.

When \(n = 2\) the brackets are \(1\) and \(9\). The first bracket changes by \(4\) each time \(n\) goes up by 1.

So the first bracket is 1 only for this one value of n and the second bracket is not prime, so no term is prime.

The answer is No.

Expand \((a - b)(a + b)\):

\((a - b)(a + b) = a^2 + ab - ab - b^2 = a^2 - b^2\)

So \(a^2 - b^2 = (a - b)(a + b)\)

Let \(a = 2^{ 107 }\). Since \(2^{ 214 } = (2^{ 107 })^2\), \(a^2 = 2^{ 214 }\), so

\(2^{ 214 } - 1 = a^2 - 1\)

Using part (a) with \(b = 1\):

\(a^2 - 1 = (a - 1)(a + 1) = (2^{ 107 } - 1)(2^{ 107 } + 1)\)

\(a = 2^{ 107 }\) is even, so \(a - 1\) and \(a + 1\) are both odd, and they differ by \(2\).

So \(2^{ 214 } - 1\) factorises as the product of two consecutive odd numbers.

Both terms divide by \(2\), so take it out first.

\(98t^2 - 8 = 2(49t^2 - 4)\)

What is left is a difference of two squares, because \(49t^2 = (7t)^2\) and \(4 = 2^2\).

Using \(A^2 - B^2 = (A + B)(A - B)\) with \(A = 7t\) and \(B = 2\),

\(49t^2 - 4 = (7t + 2)(7t - 2)\)

So altogether

\(2(7t + 2)(7t - 2)\)

Multiplying the brackets out has to give back the left-hand side, so compare a term you can see all of.

The \(v^2\) term comes from \(5v\) times \(gv\) and nothing else.

\(5g = 25\)

\(g = 25 \div 5 = 5\)

The constant term comes from \(-8\) times \(h\) and nothing else.

\((-8) \times h = 48\)

\(h = 48 \div (-8) = -6\)

Check it against the \(v\) term as well: \(5 \times (-6) + 5 \times (-8) = -70\)

Multiply the coefficient of \(m^2\) by the constant term.

\(18 \times 5 = 90\)

Now find two numbers that multiply to give \(90\) and add to give \(-33\). They are \(-30\) and \(-3\).

Use them to split the middle term.

\(18m^2 - 30m - 3m + 5\)

Take a factor out of the first two terms, then out of the last two.

\(6m(3m - 5) -(3m - 5)\)

Both parts now share the bracket \((3m - 5)\), so it comes out.

\((3m - 5)(6m - 1)\)