Factorisation

When every term contains a common number, variable, or expression, taking it out rewrites the polynomial as a product of simpler factors. Adding exponents or squaring terms is not factorisation. Exam tip: first check the HCF of all terms.

For \(x^2+5x+5\), the discriminant is \(b^2-4ac=5^2-4(1)(5)=5\), which is not a perfect square. Therefore, its roots are not rational, so it cannot be written as a product of two linear binomials with integer coefficients. In contrast, \(x^2+7x+12=(x+3)(x+4)\). Exam tip: if the discriminant of a quadratic is not a perfect square, factorisation into integer-coefficient linear factors is not possible.

The claim is correct. Using \((x-y)^2=x^2-2xy+y^2\), take \(x=7a\) and \(y=5b\); the middle term is \(-2\times7a\times5b=-70ab\). Exam tip: check the square roots of end terms, then verify \(2xy\).

In option A, \(49p^2=(7p)^2\) and \(25q^2=(5q)^2\). Therefore, \(49p^2-25q^2=(7p)^2-(5q)^2=(7p-5q)(7p+5q)\). Option B is a sum, not a difference, of perfect squares. Exam tip: before applying \(a^2-b^2=(a-b)(a+b)\), check that both terms are perfect squares.

A perfect-square trinomial has the form \(x^2-2xy+y^2=(x-y)^2\). Here, \(9a^2=(3a)^2\), \(16b^2=(4b)^2\), and the middle term is \(-24ab=-2(3a)(4b)\). Hence, \(9a^2-24ab+16b^2=(3a-4b)^2\). In option B, the last term is \(15b^2\), but it must be \(16b^2=(4b)^2\). Exam tip: take the square roots of the first and last terms and check whether the middle term is \(\pm2\) times their product.

Expanding \((x+p)(x+q)\) gives \(x^2+(p+q)x+pq\). Comparing coefficients, \(p+q=m\) and \(pq=n\). Option C incorrectly uses the difference. Exam tip: match the sum with the middle coefficient and the product with the constant term.

The expression follows the perfect-square identity \(a^2+2ab+b^2=(a+b)^2\). On taking \(a=x^2\) and \(b=1\), we get \(x^4+2x^2+1=(x^2+1)^2\). The closest distractor, \((x^2-1)^2\), expands to \(x^4-2x^2+1\), so its middle term has the wrong sign. Exam tip: treat \(x^2\) as one term when identifying perfect-square identities.

Since \(25a^2=(5a)^2\) and \(49b^2=(7b)^2\), \(25a^2-49b^2=(5a-7b)(5a+7b)\). Option A is a perfect-square trinomial instead. Exam tip: first check whether both terms are perfect squares.

Here \(9x^2=(3x)^2\) and \(4y^2=(2y)^2\). The middle term is \(2\times3x\times2y=12xy\), so the trinomial is \((3x+2y)^2\). In B, the negative last term does not fit a perfect-square trinomial. Exam tip: check the middle term using twice the product of the square roots.

First take (3y) common and inside (x^2-4x+4=(x-2)^2). Exam tip: identify the inner trinomial too.

First take (5) common and split (x^2-9) by difference of squares. Exam tip: look further after common factor.

\(m^2+16=m^2+4^2\) is a sum of two squares, so it has no factorisation into non-constant polynomials with integer coefficients. The other options use \(A^2-B^2=(A-B)(A+B)\). Exam tip: distinguish a sum from a difference of squares.

(-5+(-6)=-11) and ((-5)(-6)=30). Exam tip: for a positive last term use same-sign factors.

For a perfect-square trinomial, the square of the middle coefficient equals four times the product of the first and last coefficients. Here, \(14^2=4\times1\times49=196\), so it has identical binomial factors. Exam tip: check \(b^2=4ac\).

In a perfect-square trinomial, the middle term must match \(2abx\) and the constant must be \(b^2\). Here, \(-10x=2(x)(-5)\) and \(25=(-5)^2\), so \(x^2-10x+25=(x-5)^2\). In option B, 20 is not a perfect square. Exam tip: check the square root of the constant against the middle term.

For \(x^2+5x+7\), two integers must have product 7 and sum 5. The pair 1 and 7 has sum 8, so such factors do not exist. In contrast, \(x^2+5x+6=(x+2)(x+3)\). Exam tip: check the product first, then the sum.

\(x^2+x-12=(x+4)(x-3)\), so the constants \(+4\) and \(-3\) have opposite signs. Their product is \(-12\). In option B, the constants are both positive. Exam tip: a negative constant product indicates opposite signs.

For 6x² + 11x + 3, split 11x into 9x + 2x: 3x(2x + 3) + 1(2x + 3). Thus it factorises as (3x + 1)(2x + 3). The other choices have no suitable integer pair. Exam tip: first check the product ac.

On expanding, \(m+n=p\) and \(mn=q\). Since \(q<0\), the integers have opposite signs; since \(p>0\), the positive integer has greater magnitude. Exam tip: check both sum and product.

On multiplying we get (6x^2+8x+9x+12=6x^2+17x+12). Exam tip: check by adding the middle terms.