Factorise a quadratic expression completely

Which product has both the required middle term and constant?

These formal algebra tiles show the collected coefficients when two monic linear factors are multiplied. Filled tiles are positive and hollow tiles negative; they do not represent physical lengths. Change both constants to obtain a quadratic with middle coefficient 1 and constant −6.

Which pair gives middle coefficient 1 and constant −6?

Two conditions, one factorisation

The two cross-products combine to make the middle term. Multiplication is commutative, so the brackets can swap. A table with separate coefficient boxes needs a stated factor order

1. 1Monic means leading coefficient 1 In the expansion, m + n is the middle coefficient and mn is the constant. Match both to the original trinomial \(\displaystyle \begin{gathered}(x+m)(x+n)\\=x^2+(m+n)x+mn\end{gathered}\)
2. 2Use both conditions The sign and size of the middle coefficient and the constant must both match
3. 3Retain a common factor If every coefficient has a common factor, remove it first and retain it outside the final brackets
4. 4Multiply back Expand all four products. Collecting like terms must restore the original polynomial

What product must the two constants have?

What sum must the two constants have?

The constants have product 12 and sum 7. Enter the smaller constant

Target factorisation

\(\displaystyle \begin{aligned}&x^2+7x+12\\&=(x+3)(x+?)\end{aligned}\)

Enter the other constant

Which expansion checks every term?

A monic trinomial has constant 15 and middle coefficient −8. What signs must the constants in its two monic linear factors have?

Find constants with product 15 and sum −8. Enter the smaller signed constant

Which product restores both signs?

What product must the constants have?

The constants have sum 1 and product −6. What is the positive constant?

Which repeated factor gives the complete trinomial?

What is the highest positive numerical common factor?

Which complete factorisation reproduces the expression and retains the common factor?

Can this quadratic be written as two monic linear brackets with integer constants?

Complete the square: what must be subtracted to preserve the original expression?

A second route to the same factors

Completing the square can expose a difference of squares. Keep the expression equivalent at every step

1. 1Complete the square Half the middle coefficient is 3. Add and subtract 9. \(\displaystyle x^2+6x+5=(x+3)^2-4\)
2. 2Use a square difference Treat x + 3 as one quantity and 4 as 2² \(\displaystyle (x+3)^2-2^2=(x+1)(x+5)\)

Which factors follow from this square difference?

A leading coefficient other than 1

Simple non-monic factorisation is part of Year 10. For ax² + bx + c, split the middle term using integers with product ac and sum b, then group

1. 1Compute ac For 6x² + 7x −3, the required product is 6 × (−3) = −18.
2. 2Find the split −2 and 9 have product −18 and sum 7.
3. 3Group with matching brackets 6x² −2x + 9x −3 becomes 2x(3x −1) + 3(3x −1)
4. 4Check the product (2x + 3)(3x −1) expands back to 6x² + 7x −3.

What is the required product ac?

Split the middle coefficient 7 into two integers with product −18. Enter the smaller integer

Required equality

\(\displaystyle \begin{aligned}&6x^2+7x-3\\&=6x^2-2x+\Box x-3\end{aligned}\)

What is the other split coefficient?

Which product expands to the original trinomial?

Which line extracts the matching bracket after splitting the middle term?

Is this product wrong, or merely unfinished?