Question

How do you simplify the product \[(x + 7)(x + 5)\] and write it in standard form?

Answer 1

Hint: We simplify or solve the product given by multiplying numbers in the first bracket with the complete second bracket one by one. We multiply the given factors or brackets by opening the terms and then write the quadratic equation formed by multiplication in a way that it matches general quadratic equation i.e. \[a{x^2} + bx + c\]
* When we have to multiply the brackets like \[(a + b)(c + d)\] then we multiply the second bracket by a first and then by b i.e. \[a(c + d) + b(c + d)\]. Then we multiply each term outside the bracket with terms inside the bracket one by one i.e. a with c, a with d, b with c and b with d i.e. \[ac + ad + bc + bd\]

Complete step-by-step answer:
We are given the product \[(x + 7)(x + 5)\]
We will simplify the product by multiplying terms from first bracket to second bracket one by one.
\[ \Rightarrow (x + 7)(x + 5) = x(x + 5) + 7(x + 5)\]
Now we multiply each term outside the bracket with terms inside the bracket one by one
\[ \Rightarrow (x + 7)(x + 5) = \left( {x \times x} \right) + \left( {x \times 5} \right) + \left( {7 \times x} \right) + \left( {7 \times 5} \right)\]
Calculate the products on right side of the equation
\[ \Rightarrow (x + 7)(x + 5) = {x^2} + 5x + 7x + 35\]
Add the terms having same coefficient or variable associated with them
\[ \Rightarrow (x + 7)(x + 5) = {x^2} + 12x + 35\]
Now we have right hand side of the equation as a quadratic equation.
Since standard form of a general quadratic equation is \[a{x^2} + bx + c\]

\[\therefore \]Standard form of the quadratic equation formed by the product \[(x + 7)(x + 5)\] is \[{x^2} + 12x + 35\]

Note:
Many students make the mistake of writing the quadratic equation formed without simplifying i.e. without even adding the terms or combining the terms with the same coefficient or variable associated with it. Keep in mind the general quadratic equation has only b along with x so we have to write combined or complete value of b.