Find the factors of $x^2+9x+20$ and then compare it with $x^2+(a+b)x+ab$.
Given :
Given equation is $x^2+9x+20.$
To do :
We have to factorise it and compare with $x^2+(a+b)x+ab$
Solution :
To factorise it we have to find two numbers whose sum is equal to the coefficient of x and the product is equal to the constant term.
$9x=(4+5)x$ [$4\times5=20$]
Therefore,
$x^2+9x+20$=$x^2+(4+5)x+4\times5$
Comparing it with $x^2+(a+b)x+ab$, we get,
a=4 and b=5.
$x^2+9x+20$=$x^2+(4+5)x+4\times5$
=$x^2+4x+5x+4\times5$
=$x(x+4)+5(x+4)$
=$(x+4)(x+5)$
The factors of $x^2+9x+20$ are $(x+4)$ and $(x+5)$.
Related Articles
- If \( 2 x+3 \) and \( x+2 \) are the factors of the polynomial \( g(x)=2 x^{3}+a x^{2}+27 x+b \), find the value of the constants $a$ and $b$.
- Factorize the expression $ab(x^2+1)+x(a^2+b^2)$.
- $( x-2)$ is a common factor of $x^{3}-4 x^{2}+a x+b$ and $x^{3}-a x^{2}+b x+8$, then the values of $a$ and $b$ are respectively.
- Solve by any method except cross multiplication$a(x+y)+b(x-y)=a^2-ab+b^2$ and $a(x+y)-b(x-y)=a^2+ab+b^2$
- If \( x+1 \) is a factor of \( 2 x^{3}+a x^{2}+2 b x+1 \), then find the values of \( a \) and \( b \) given that \( 2 a-3 b=4 \).
- Find $\alpha$ and $\beta$, if $x + 1$ and $x + 2$ are factors of $x^3 + 3x^2 - 2 \alpha x + \beta$.
- If $x + \frac{1}{x} =20$, find the value of $x^2 + \frac{1}{x^2}$.
- $x^2+\frac{1}{x^2}=34$, then find the value of $x+\frac{1}{x}$.
- If both $x + 1$ and $x - 1$ are factors of $ax^3 + x^2 - 2x + b$, find the values of $a$ and $b$.
- Which of the following is a factor of $f( x)=x^{2}-9 x+20?$$A).\ ( x-2)$$B).\ ( x-3)$$C).\ ( x-4)$$D).\ ( x-5)$
- If $p(x) = x^2 - 2\sqrt{2}x+1$, then find the value of $p(2\sqrt{2})$.
- If $x+\frac{1}{x}=4$, then find, $x^{2}+\frac{1}{x^2}=?$
- If $x^2-6x+1=0$, then find the value of $x^2+\frac{1}{x^2}$.
- Find the value of $x$:$\frac{7}{2} x-\frac{5}{2} x=\frac{20 x}{3}+10$.
- If $2x + 3$ and $x + 2$ are the factors of the polynomial $g(x) = 2x^3 + ax^2 + 27x + b$, then find the values of the constants a and b in the polynomial g(x).
Kickstart Your Career
Get certified by completing the course
Get Started