UPSKILL MATH PLUS
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Learn moreProduct of three binomials:
Consider the product of the binomials \((x+a)(x+b)(x+c)\).
Apply the distributive property for the first two terms as follows:
\([(x+a)(x+b)](x+c)\) \(=\) \([(x\times x)]\)\(+(x\times b)\)\(+(a\times x)\)\(+(a\times b)]\)\((x+c)\)
\(=\) \((x^2+bx+ax+ab)\)\((x+c)\)
Again apply distributive law.
\((x^2+bx+ax+ab)(x+c)\)
\(=\) \((x^2\times x)\)\(+(bx\times x)\)\(+(ax\times x)\)\(+(ab\times x)\)\(+(x^2\times c)\)\(+(bx\times c)\)\(+(ax\times c)\)\(+(ab\times c)\).
\(=\) \(x^3+bx^2\)\(+ax^2+abx\)\(+cx^2+bcx\)\(+acx+abc\)
Separate the cubic, square, variables and constants terms.
\(=\) \(x^3+ax^2\)\(+bx^2+cx^2\)\(+abx+bcx\)\(+acx+abc\)
\(=\) \(x^3+(a+b+c)x^2+(ab+bc+ac)x+abc\)
Thus we have the identity \((x+a)(x+b)(x+c)\) \(=\) \(x^3\)\(+(a+b+c)x^2\)\(+(ab+bc+ca)x+abc\).
Example:
\((4y+5)(4y+3)(4y-7)\)
Let us use the identity, \((x+a)(x+b)(x+c)\) \(=\) \(x^3+(a+b+c)x^2\)\(+(ab+bc+ca)x\)\(+abc\)
Comparing \((4y+5)(4y+3)(4y-7)\) with \((x+a)(x+b)(x+c)\), we have \(x=4y, a=5, b=3\) an d\(c=-7\).
Substitute the known values.
\((4y+5)(4y+3)(4y-7)\) \(=\) \((4y)^3\)\(+(5+3-7)(4y)^2\)\(+((5\times 3) + (3\times -7) +(-7\times 5))(4y)\)\(+5 \times 3 \times -7)\)
\((4y+5)(4y+3)(4y-7)\) \(=\) \(64y^3+16y^2\)\(+(15-21-35)(4y)-105\)
\((4y+5)(4y+3)(4y-7)\) \(=\) \(64y^3+16y^2\)\(-164y-105\)
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