Answer:
The roots of [tex]\bold{\left(x^{2}+3 x-4\right)\left(x^{2}-4 x+29\right)}[/tex] is -4 and 1
Solution:
From question, given that [tex]\left(x^{2}+3 x-4\right)\left(x^{2}-4 x+29\right)[/tex]
To factorize the first expression that is [tex]\left(x^{2}+3 x-4\right)[/tex] ,follow the below steps:
[tex]\left(x^{2}+3 x-4\right)[/tex] ---- eqn (1)
“3x” can be rewritten as 4x-x. Now the above equation becomes,
[tex]=\left(x^{2}+4 x-x-4\right)[/tex]
By taking x as common from [tex]x^{2}+4 x \text { and }[/tex] -1 from -x-4 the above equation becomes,
=x (x+4) - 1(x+4)
=(x+4)(x-1)
Hence the factors of [tex]\left(x^{2}+3 x-4\right) \text { is }(x+4)(x-1)[/tex]
so the roots of [tex]\left(x^{2}+3 x-4\right)[/tex] is -4 and 1.
Now to factorize the second expression that is [tex]\left(x^{2}-4 x+29\right)[/tex] ,follow the below steps:
[tex]\left(x^{2}+3 x-4\right)[/tex] --- eqn (2)
Equation (2) cannot be factorized (imaginary roots)
Hence the roots of [tex]\left(x^{2}+3 x-4\right)\left(x^{2}-4 x+29\right)[/tex] is -4 and 1.
