Multiplying Polynomials - Quadratics

Given \(2\) linear expressions \(ax+b\) and \(cx+d\), where \(a,b,c\) and \(d\) are real numbers and \(a,c \neq 0\), we have

\[\begin{align} (ax+b)(cx+d) &=ax(cx+d)+b(cx+d)\\ &=ax(cx)+ax(d)+b(cx)+b(d)\\ &=acx^2+adx+bcx+bd\\ &=(ac)x^2+(ad+bc)x+bd. \end{align}\]

To find the sum of the coefficients in the expansion of \((ax+b)(cx+d)\), we can simply substitute \(x=1\). In other words, the sum of the coefficients in the expansion of \((ax+b)(cx+d)\) is

\[(a\cdot 1+b)(c\cdot 1+d)=(a + b)(c + d).\]

Find the sum of the coefficients in the expansion of \((x+2)(3x+4)\).

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Solution \(1\):
By expanding, we have \((x+2)(3x+4)=3x^2+10x+8\). Hence, the sum of the coefficients is \(3+10+8=21. \ _\square\)

Solution \(2\):
By the substitution of \(x=1\), the sum of the coefficients is \[(1 \cdot 1 + 2)(3 \cdot 1 + 4) = (1+2)(3+4) = 3 \cdot 7 = 21. \ _\square\]

Given \(2\) quadratics \(ax^2+bx+c\) and \(dx^2+ex+f\), where \(a,b,c,d,e\) and \(f\) are real numbers and \(a, d \neq 0\), we have

\[\begin{align} &\left(ax^2+bx+c)(dx^2+ex+f\right)\\ &=ax^2\left(dx^2+ex+f\right)+bx\left(dx^2+ex+f\right)+c\left(dx^2+ex+f\right)\\ &=ax^2\left(dx^2\right)+ax^2(ex)+ax^2(f)+bx\left(dx^2\right)+bx(ex)+bx(f)+c\left(dx^2\right)+c(ex)+c(f)\\ &=adx^4+aex^3+afx^2+bdx^3+bex^2+bfx+cdx^2+cex+cf\\ &=adx^4+aex^3+bdx^3+afx^2+bex^2+cdx^2+bfx+cex+cf\\ &=(ad)x^4+(ae+bd)x^3+(af+be+cd)x^2+(bf+ce)x+cf. \end{align}\]

To find the sum of the coefficients in the expansion of \((ax^2+bx+c)(dx^2+ex+f)\), we can simply substitute \(x=1\):

\[(a \cdot 1^2 + b \cdot 1 + c)(d \cdot 1^2 + e \cdot 1 + f)=(a + b + c)(d + e + f).\]

Find the sum of the coefficients in the expansion of \((x^2+2x+3)(4x^2+5x+6)\).

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Solution \(1\):
By expanding, we have \((x^2+2x+3)(4x^2+5x+6)=4x^4+13x^3+28x^2+27x+18\). Hence, the sum of the coefficients is \(4+13+28+27+18=90. \ _\square\)

Solution \(2\):
By the substitution \(x=1\), the sum of the coefficients is \[(1^2 + 2 \cdot 1 + 3)(4 \cdot 1^2 + 5\cdot 1 + 6) = (1+2+3)(4+5+6) = 6 \cdot 15 =90. \ _\square\]

Another way to multiply polynomials is to use the long multiplication technique. It is like multiplying two numbers, the only difference being that it is with variables. Look through the following example:

[[start-example]]

Step 1: Align the equation in this format.

Step 2: Multiply the equation at the top with 5 like this:

Step 3: Multiply the top equation with the number left of five (which is \(2x\)) to get the answer in the purple box.

Note: The numbers that have the same variables are aligned vertically, and the ones that are not are left aside for the time being, which are still needed!

Step 4: Multiply the top equation with the number left of 2x(which is x raise to power 2) to get the answer in the red box.

Step 5: Add the white box, the purple box and the red box to get a final answer, which is the green box. This is arranged in degree order (highest powers to lowest) in the pink box below, and there you go. You're done!